Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg

Which Of These Boxes Will Not Accelerate!30 Newtons40 Newtons50 Kg15 NewtonB.10 Kg30 NewtonsC.30 Newtons80

Answers

Answer 1

Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.


Related Questions

During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.

Answers

Answer:

the angular speed of the person increases, being able to make more turns and faster.

 K₂ = K₁ I₁ / I₂

Explanation:

When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved

initial, joint when starting to turn

         L₀ = I₁ w₁

final. When you shrink your arms and legs

         Lf = I₂ w₂

         L₀ = Lf

         I₁ w₁ = I₂ w₂

when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases

      I₂ <I₁

         w₂ = I₁ / I₂ w₁

therefore the angular speed of the person increases, being able to make more turns and faster.

When it goes into the water it straightens the arm and leg, so the moment of inertia increases

          I₁> I₂

           w₁ = I₂ / I₁ w₂

therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.

In both cases the kinetic energy is

          K = ½ I w²

the initial kinetic energy is

          K₁ = ½ I₁ w₁²

the final kinetic energy is

          K₂ = ½ I₂ w₂²

we substitute

          K₂ = ½ I₂ (I₁ / I₂ w1² 2

          K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²)  I₁ / I₂  

          K₂ = K₁ I₁ / I₂

therefore we see that the kinetic energy increases by factor I₁/I₂

Which of the following best describes the current age of the Sun?

A.) It is near the end of its lifespan.

B.) It is about halfway through its lifespan.

C.) It is early in its lifespan.

D.) We do not have a good understanding of the Sun's age.

Answers

Answer:  Its b, The only problem with this is is there supposed to be a picture?

Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.

Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.

Answers

Answer:

The velocity is  [tex]v_2 = 6.8 \ m/s[/tex]

The pressure is  [tex]P_2 = 204978 Pa[/tex]

Explanation:

From the question we are told that

 The speed at which water is travelling through is  [tex]v = 1.7 \ m/s[/tex]

  The pressure is  [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]

   The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]

Where D is the diameter of first pipe

   

According to the principal of continuity we have that

       [tex]A_1 v_1 = A_2 v_2[/tex]    

Now  [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as

       [tex]A_1 = \pi \frac{D^2}{4}[/tex]

and  [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as  

       [tex]A_2 = \pi \frac{d^2}{4}[/tex]

Recall   [tex]d = \frac{D}{2}[/tex]

        [tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]

        [tex]A_2 = \frac{A_1}{4}[/tex]

So    [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]

substituting value

        [tex]1.7 = \frac{1}{4} * v_2[/tex]    

        [tex]v_2 = 4 * 1.7[/tex]    

       [tex]v_2 = 6.8 \ m/s[/tex]

   

According to Bernoulli's equation  we have that

     [tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]

substituting values

     [tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]

     [tex]P_2 = 204978 Pa[/tex]

28 points!! please help

Answers

7(a) transpiration is faster in warmer dry air
(b)(I)xylem
(ii) 1. They are stacked end-to-end
2. Consists of dead cells

A solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins

Answers

Answer:

Cube temperature = 526.83 K

Explanation:

Volume of the cube and sphere will be the same.

Now, volume of cube = a³

And ,volume of sphere = (4/3)πr³

Thus;

a³ = (4/3)πr³

a³ = 4.1187r³

Taking cube root of both sides gives;

a = 1.6119r

Formula for surface area of sphere is;

As = 4πr²

Also,formula for surface area of cube is; Ac = 6a²

Thus, since a = 1.6119r,

Then, Ac = 6(1.6119r)²

Ac = 15.5893r²

The formula for radiant power is;

Q' = eσT⁴A

Where;

e is emissivity

σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k

T is temperate in kelvin

A is Area

So, for the cube;

(Qc)' = eσ(Tc)⁴(Ac)

For the sphere;

(Qs)' = eσ(Ts)⁴(As)

We are told (Qc)' = (Qs)'

Thus;

eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)

eσ will cancel out to give;

(Tc)⁴(Ac) = (Ts)⁴(As)

Since we want to find the cube's temperature Tc,

(Tc)⁴ = [(Ts)⁴(As)]/Ac

Plugging in relevant figures, we have;

(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²

r² will cancel out to give;

(Tc)⁴ = [556⁴ × 4π]/15.5893

Tc = ∜([556⁴ × 4π]/15.5893)

Tc = 526.83 K

Thana reminds Alston that because the electric field is uniform, a constant electric force is exerted on the electron. Alston recognizes that, in this case, they can use the kinematic equations to describe the motion of the charged particle while it is inside the region containing the electric field. He asks Thana to write down an equation they can use to calculate the acceleration of the particle while it is inside the region containing a uniform electric field. Which of These equations is correct?

Answers

Answer:

  a = - e E / m

a = - 1,758 10¹¹ E

Explanation:

For this exercise we can use Newton's second law

        F = m a

where the force is electric

 the forces given by the product of the charge by the electric field

         F = q E

in this case tell us that the charge is the charge of the electron

         q = -e = - 1.6 10⁻¹⁹ C

we substitute

        - e E = m a

          a = - e E / m

we calculate

           a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E

           a = - 1,758 10¹¹ E

The negative sign indicates that the acceleration is in the opposite direction to the electric field

Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosmonaut Valdimir tosses a banana at you at a speed of 16 m/s. At exactly the same instant, you fling a scoop of ice cream at Valdimir along exactly the same path. The collision between banana and ice cream produces a banana split 8.2 m from your location 1.4 s after the banana and ice cream were launched.

1. How fast did you toss the ice cream?

2. How far were you from Valdimir when you tossed the ice cream?

Answers

Answer:

a

The speed is   [tex]s = 5.857 m/s[/tex]

b

The distance is  [tex]D = 22.4 \ m[/tex]

Explanation:

From the question we are told that

     The speed of the banana is  [tex]v = 16 \ m/s[/tex]

   The distance from my  location is  [tex]d = 8.2 \ m[/tex]  

     The time taken is  [tex]t = 1.4 \ s[/tex]

The speed of the ice cream is

          [tex]s = \frac{d}{t}[/tex]

substituting values

        [tex]s = \frac{8.4}{1.4}[/tex]

        [tex]s = 5.857 m/s[/tex]

The distance of separation between i and Valdimir is the same as the distance covered by the banana

   So  

          [tex]D = v * t[/tex]

substituting values

        [tex]D = 16 * 1.4[/tex]

        [tex]D = 22.4 \ m[/tex]

     

Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.

Answers

Answer:

Therefore, the distance between politician and TV set is 2536km

Explanation:

Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.

The reporter hears the sound is

4.1 / 343 = 0.01195 s later

The viewer hears the sound from the TV is

2.9 / 343 = 0.00845s

the difference is 0.00845 sec

the question is how far the TV signal can travel in that time.

the distance between politician and TV set is

= 0.00845 * 3*10^8 m

= 2536 km

d = 2536km

Therefore, the distance between politician and TV set is 2536km

To get up on the roof, a person (mass 69.0 kg) places a 6.40 m aluminum ladder (mass 11.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom

Answers

Answer:

N = 243.596 N ≈ 243.6 N

Explanation:

mass of person = 69 kg ( M )

mass of aluminium ladder = 11 kg ( m )

length of ladder = 6.4 m ( l )

base of ladder = 2 m from the house (d )

center of mass of ladder = 2 m from the bottom of ladder

person on ladder standing = 3 m from bottom of ladder

Calculate the magnitudes of the forces at the top and bottom of the ladder

The net torque on the ladder = o ( since it is at equilibrium )

assuming: the weight of the person( mg) acting at a distance x along the ladder. the weight of the ladder ( mg) acting halfway along the ladder and the reaction N acting on top of the ladder

X = l/2

x = 6.4 / 2 = 3.2

find angle formed by the ladder

cos ∅ = d/l

    ∅ = [tex]cos^{-1][/tex] 2/6.4 = [tex]cos^{-1}[/tex]0.3125  ≈ 71.79⁰

remember the net torque around is = zero

to calculate the magnitude of forces on the ladder we apply the following formula

[tex]N = \frac{mg(dcosteta)+ Mgxcosteta}{lsinteta}[/tex]

m = 11 kg, M = 69 kg, l = 6.4 , x = 3,  teta( ∅ )= 71.79⁰, g = 9.8

back to equation  N = [tex]\frac{11*9.8(2*cos71.79)+ 69*9.8*3* cos71.79}{6.4sin71.79}[/tex]

N = (67.375 + 633.938) / 2.879

N = 243.596 N ≈ 243.6 N

1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.

Answers

Explanation:

It is given that,

Mass of a car is 900 kg

Radius of curve is 600 m

Speed of the car in the curve is 25 m/s

We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :

[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]

So, the centripetal force exerted on a car is 937.5 N.

Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.  

A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?

Answers

Answer:

The width of the river is  [tex]z = 60.62 \ m[/tex]

Explanation:

From the question we are told that

     The distance of the base line is D = 130 m

       The angle is  [tex]\theta = 25^o[/tex]

A diagram illustration the question is shown on the first uploaded image

    Applying Trigonometric Rules for Right-angled Triangle,

            [tex]tan 25 = \frac{z}{130}[/tex]

Now making  z the subject

           [tex]z = 130 * tan (25)[/tex]

          [tex]z = 60.62 \ m[/tex]

In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]

Answers

Answer:

Explanation:

                                                                           

The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]

What is equivalent resistance?

The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.

What is equivalent resistance in series?

Resistors are in series whenever the current flows through the resistors sequentially. It is given by

[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]

What is equivalent resistance in parallel?

Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.

The equivalent resistance is the total resistance measured in a parallel. It is given by

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]

Given:

Resistor, [tex]R_{1} = 100 ohm[/tex]

Resistor, [tex]R_{2} = 200 ohm[/tex]

Voltage, [tex]V = 10 Volt[/tex]

Since, resistors are connected in parallel, the equivalent resistor is given by,

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]

[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]

[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]

Hence, the equivalent resistor is [tex]66.66 ohm[/tex].

To learn more about equivalent resistor here

https://brainly.com/question/113987

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If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

- 2x
- 1/4
- 1/2
- 4x

Answers

Answer:

The correct option is;

- 4x

Explanation:

From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source

The inverse square law can be presented as follows;

[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]

As the distance, r, increases, the surface it covers also increases by the power of 2

Therefore, where the distance increases from r to 2·r, we have;

When, I, remain constant

[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]

The surface increases to 4·S by the inverse square law

Therefore, the correct option is 4 × x.

A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.

What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.

Answers

Answer:

J = 0.800 kg m/s

Fmax = 291 N

Explanation:

During the fall, energy is conserved.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.81 m/s² × 1.45 m)

v = 5.33 m/s

Alternatively, you can use kinematics to find the velocity.

Impulse = change in momentum

J = Δp

J = mΔv

J = (0.150 kg) (5.33 m/s)

J = 0.800 kg m/s

Impulse = area under F vs t graph

J = ∫ F dt

J = ½ Fmax Δt

(0.800 kg m/s) = ½ Fmax (0.0055 ms)

Fmax = 291 N

1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain ​

Answers

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

I really need help with this question someone plz help !

Answers

Answer:weight

Explanation:weight

A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s

Answers

Answer:

a) 98 m/s

b) 19.6 s

c)  1449.8 m

d)  29.6 s

e)  168.6 m/s

f)  46.8 s

Explanation:

Given that

Acceleration of the rocket, a = 5 m/s²

Altitude of the rocket, s = 960 m

a)

Using the equation of motion

v² = u² + 2as, considering that the initial velocity, u is 0. Then

v² = 2as

v = √2as

v = √(2 * 5 * 960)

v = √9600

v = 98 m/s

b)

Using the equation of motion

S = ut + ½at², considering that initial velocity, u = 0. So that

S = ½at²

t² = 2s/a

t² = (2 * 960) / 5

t² = 1920 / 5

t² = 384

t = √384 = 19.6 s

c)

Using the equation of motion

v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that

0 = 98² + 2(-9.8) * s

9600 = 19.6s

s = 9600/19.6

s = 489.8 m

The maximum altitude now is

960 m + 489.8 m = 1449.8 m

d)

Using the equation of motion

v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that

0 = 98 +(-9.8 * t)

98 = 9.8t

t = 98/9.8

t = 10 s

Total time then is, 10 + 19.6 = 29.6 s

e) using the equation of motion

v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,

v² = 0 + 2 * 9.8 * 1449.8

v² = 28416.08

v = √28416.08

v = 168.6 m/s

f) using the equation of motion

S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s

1449.8 = 0 + ½ * 9.8 * t²

2899.6 = 9.8t²

t² = 2899.6/9.8

t² = 295.88

t = √295.88

t = 17.2 s

total time in air then is, 17.2 + 29.6 = 46.8 s

PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work

Answers

B

HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION

Mr. Patel is photocopying lab sheets for his first period class. A particle of toner carrying a charge of 4.0 * 10^9 C in the copying machine experiences an electric field of 1.2 * 10^6 N/C as it’s pulled toward the paper. What is the electric force acting on the toner particle?

Answers

Answer:

4.8 × 10^15 N

Explanation:

Electric Field is defined as Force per unit Charge.

This is expressed mathematically as;

E= F/Q

Where E- Electric Field

F- Force

Q- charge

From the expression above by change of subject of formula for F, we have;

F=E×Q

= 1.2 * 10^6 ×4.0 * 10^9

= 4.8 × 10^15 N

A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units

Answers

Answer:

Final Temperature = 71 °C

Explanation:

In this case, the ideal gas equation is written as;

PV = mRT

Where;

P is pressure

V is volume

m is mass

R is gas constant

T is temperature

We will take the volume to be constant.

So, in the initial state, we have;

P1•V = m1•R•T1 - - - eq(1)

In the final state, we have;

P2•V = m2•R•T2 - - - - eq(2)

Combining eq (1) and eq(2),we have;

P1•m2•R•T2 = P2•m1•R•T1

Dividing both sides by R gives;

P1•m2•T2 = P2•m1•T1

Making T2 the subject gives;

T2 = (P2•m1•T1)/(P1•m2)

Now, we are given;

m1 = 2kg

m2 = ½*2 = 1kg

P1 = 4 atm

P2 = 2.2 atm

T1 = 40°C = 273 + 40 K = 313K

Plugging in this values into the T2 equation, we have;

T2 = (2.2 × 2 × 313)/(4 × 1)

T2 = 344 K

Converting to °C, we have;

T2 = 344 - 273 = 71 °C

The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?

Answers

Answer:

61.6° west of South

Explanation:

The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.

Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side  and the hypotenuse is 8

cos(θ) = [adjacent/hypotenuse]

Cos θ = 3.8/8

Cos θ = 0.475

θ = cos^-1 (0.475)

θ = 61.6°

Therefore the angle is 61.6° west of South.

Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=

Answers

Answer:

Explanation shown below.

Explanation:

1.The number of images formed by 2 parallel mirrors is an infinite number of images.

2. The characteristics of a plane mirror is such that the object distance equals the image distance.

Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.

Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.

Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .

Hence the shortest distances are 8.7cm and 68.9cm

3. The number of reflections is infinite for both cases.

A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is

Answers

Answer:

v = 1.4 m/s

Explanation:

This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):

[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex]  (1)

m1: mass of the toy car = 2 kg

m2: mass of the toy truck = 3 kg

v1: speed of the toy car = 1 m/s

v2: speed of the truck car = 3 m/s

v: speed of both car and truck after the collision = ?

In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.

You solve the equation (1) for v, and you replace the values of all variables involved:

[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]

this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck

Hence, the speed of both car and truck toys is 1.4 m/s

Someone plzzz helpppppp with this last question

Answers

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?

Answers

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na

Answers

Answer:

Explanation:

During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy

mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .

mgr  = 1/2 x ( 1/2 m r²+ mr²) x ω²

gr  = 1/2 x 1/2  r² x ω² + 1/2r² x ω²

g = 1 / 4 x ω² r + 1 / 2 x ω² r

g = 3  x ω² r/ 4

ω² = 4g /3 r

= 4 x 9.8 /  3 x  .25

= 52.26

ω = 7.23  rad / s .

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temperature of 322 K. Over the course of one hour, the engine absorbs 1.37 x 105 J from the hot reservoir and exhausts 7.4 x 104 J into the cold reservoir.

1) What is the power output of this engine?

2) What is the maximum (Carnot) efficiency of a heat engine running between these two reservoirs?

3) What is the actual efficiency of this engine?

Answers

Answer:

The power output of this engine is  [tex]P = 17.5 W[/tex]

The  the maximum (Carnot) efficiency is  [tex]\eta_c = 0.7424[/tex]

The  actual efficiency of this engine is  [tex]\eta _a = 0.46[/tex]

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  [tex]T_h = 1250 \ K[/tex]

      The temperature of the cold reservoir  is  [tex]T_c = 322 \ K[/tex]

     The energy absorbed from the hot reservoir is [tex]E_h = 1.37 *10^{5} \ J[/tex]

       The energy exhausts into  cold reservoir is  [tex]E_c = 7.4 *10^{4} J[/tex]

The power output is mathematically represented as

      [tex]P = \frac{W}{t}[/tex]

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      [tex]W = E_h -E_c[/tex]

substituting values

       [tex]W = 63000 J[/tex]

So

    [tex]P = \frac{63000}{3600}[/tex]

    [tex]P = 17.5 W[/tex]

The Carnot efficiency is mathematically represented as

          [tex]\eta_c = 1 - \frac{T_c}{T_h}[/tex]

         [tex]\eta_c = 1 - \frac{322}{1250}[/tex]

         [tex]\eta_c = 0.7424[/tex]

The actual efficiency is mathematically represented as

        [tex]\eta _a = \frac{W}{E_h}[/tex]

substituting values

         [tex]\eta _a = \frac{63000}{1.37*10^{5}}[/tex]

         [tex]\eta _a = 0.46[/tex]

     

1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?

Answers

Answer:

Energy Stored = 36000 J = 36 KJ

Explanation:

The power of a battery is given by the formula:

P = IV

where,

P = Power delivered by the battery

I = Current Supplied to the battery

V = Potential Difference between terminals of battery = 12 volt

Now, we multiply both sides by the time period (t):

Pt = VIt

where,

Pt = (Power)(Time) = Energy Stored = E = ?

It = Battery Current Rating = 50 A.min

Converting this to A.sec;

It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.

Answers

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

calculate the volume of marble if its diameter is 10mm​

Answers

Answer:

The volume of the marble is [tex]523.33\ mm^2[/tex].

Explanation:

Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.

The volume of spherical shaped object is given by :

[tex]V=\dfrac{4}{3}\pi r^3[/tex]

Plugging all the values, we get :

[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]

So, the volume of the marble is [tex]523.33\ mm^2[/tex].

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