1. When an object is at rest, not moving, and is crashed into by another
object...which object is experiencing a force and in what direction? Or are
both experiencing a force - what direction?

Answer:

more than 51.45 N

__________________________________________________________

We are given:

Mass of the television = 15 kg

Coefficient of Static friction = 0.35

Minimum force required to move the television:

Normal Force:

We know that the normal force is equal and opposite to the Weight of the television

Weight of the television = Mg            

[where m is the mass and g is the acceleration due to gravity]

Weight = 15 * 9.8

Weight = 147 N

Force of Friction:

We are given the coefficient of Friction = 0.35

We know that coefficient of Friction = Force of friction / Normal Force

replacing the variables

0.35 = Force of Friction / 147

Force of Friction = 147 * 0.35                     [multiplying both sides by 147]

Force of Friction = 51.45 N

Since a force of 51.45 N is will be applied opposite to the direction of application of Force, the television will only move when we apply more force than 51.45 N

Answer:

it is C

Explanation:

131.2 J and The last one on number 8

I gave the same answer and it passed.

7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.

8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Given:

A moving object is rolling on a surface that is 5 m off the ground,

The speed of the object, v = 4 m/s,

The mass of the object, m = 3.2 kg,

Calculate the kinetic energy after 10 meters as shown below,

KE = 1/2 × 4² × 3.2

KE = 25.6 J

Calculate the potential energy as shown below,

PE = 3.2 × 9.8 × 5

PE = 156.8 J

Thus, total energy = KE + PE

The total energy = 25.6 + 156.8

The total energy = 182.4 J

8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.

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Answer:

[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]

Explanation:

Coulomb's Law

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]

Where:

[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]

Substituting values:

[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]

[tex]q = 1.05\cdot 10^{-5}~c[/tex]

This charge corresponds to a number of electrons given by the elementary charge of the electron:

[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]

Thus, the charge of any of the spheres is:

[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]

[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]

Answer:

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.

Explanation:

please mark me brainlist

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s

Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

Answer:

lucky mauld mauldgomary was an british poet...

Answer:

Mutualism, commensalism, parasitism, competition, and predation.

Explanation:

mutualism- relationship between two or more organisms where both are benefited. Example-oxpecker with rhino/zebra. They eat bugs off of them which means that they are getting food, while the rhino/zebra are getting cleaned up with pest control.

commensalism- relationship between two organisms where one benefits and the other isn't benefited or harmed. EX- tree frogs use plants as protectioin.he frog is benefited, and the plant is neither harmed nor benefited. Remora fish have a disk on their heads that they use to attach themselves to larger animals for protection. The animals they attach to are neither harmed nor benefited.  

parasitism- in a relationship where an organism benefits at the expense of the other. (one is benefited while the other is harmed) ex- fleas and ticks that live on cats and dogs, tape worms that live in people and animals that eat the food which means that the people aren't getting the food or nutrition that they eat. lice, etc

competition- interaction within organisms/species in which both the organisms/species are harmed and is apart of natural selection. Examples may include two males fighting over a mate, animals competing over food, limited habitats that they are fighting over, territory, etc.

predation- the preying of one animal on another. It's where the predator hunts and eats another organism which is its prey. categorized within-(1) carnivory, (2) herbivory, (3) parasitism, and (4) mutualism. Each type of predation can by categorized based on whether or not it results in the death of the prey.ex- owls hunting mice, wolves hunting rabbits, lion hunting gazelle, etc.

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

Answer:

P_abs = 105120.2 N/m²

Explanation:

We are given;

Specific gravity of oil; ρ_oil = 0.6 g/cm³ = 600 kg/m³

Depth of water; h_w = 21 cm = 0.21 m

Depth of oil; h_o = 35 cm = 0.35 m

From tables specific gravity of water is; ρ_w = 1000 kg/m³

Thus, to get the absolute pressure at the bottom of the container, we will use the formula;

P_abs = (ρ_w × g × h_w) + (ρ_oil × g × h_oil) + P_a

Where P_a is atmospheric pressure with a standard value of 1.01 × 10^(5) N/m²

g is gravitational acceleration = 9.81 m/s²

Thus;

P_abs = (1000 × 9.81 × 0.21) + (600 × 9.81 × 0.35) + (1.01 × 10^(5))

P_abs = 105120.2 N/m²

Answer:

Silver

Explanation:

Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.

In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂

Answer:

x = 1127 [m]

Explanation:

In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.

[tex]v_{f} =v_{o} -a*t[/tex]

where:

Vf = final velocity = 9 [m/s]

Vo = initial velocity = 55 [m/s]

a = acceleration o desacceleration [m/s²]

t = time = 49 [s]

Now replacing:

9 = 55 - a*49

a*49 = 55 + 9

a = 1.306 [m/s²]

Note: The negative sign in the above equation means that the speed decreases.

Now using the second equation.

[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]

(9)² = (55)² - 2*(1.306)*x

2944 = 2.612*x

x = 1127 [m]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]                              

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s [/tex]

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s [/tex]      

3) When the object is at the equilibrium position, the speed of the object is:

[tex] \frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2} [/tex]    

[tex] \frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2} [/tex]      

[tex] v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s [/tex]

I hope it helps you!                                                                                        

(1) the speed of the object when compression of the spring is 8 cm is 0.449 m/s

(2) the speed of the object when compression of the spring is 5 cm is 0.648 m/s

(3) the speed of the object when the spring is at equilibrium is 0.748 m/s

Given that the mass of the object, m = 5 kg

spring constant, k = 280 N/m

compression of the spring , x = 10 cm = 0.1m

(i) the spring compression is at d = 8cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.08)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.2016[/tex]

v = 0.449 m/s

(ii) the spring compression is at d = 5cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0.05)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.42[/tex]

v = 0.648 m/s

(iii) the spring is at equilibrium so compression is at d = 0cm

according to the conservation of energy:

[tex]\frac{1}{2}kx^2=\frac{1}{2}kd^2+\frac{1}{2}mv^2[/tex]

where v is the speed at the given compression of the spring.

[tex]\frac{1}{2}\times280\times(0.1)^2=\frac{1}{2}\times280\times(0)^2+\frac{1}{2}\times5\times v^2\\\\v^2=0.56[/tex]

v = 0.748 m/s

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Answer:

V₀ₓ = 9.2 m/s

Nearest answer:

D) 8.9 m/s

Explanation:

First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:

h = V₀y t + (1/2)gt²

where,

h = height of building = 10.4 m

V₀y = vertical component of initial speed = 0 m/s

t = time = ?

g = 9.8 m/s²

Therefore,

10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (10.4 m)(2)/(9.8 m/s²)

t = √[2.122 s²]

t = 1.45 s

Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance between building and car = 13.4 m

V₀ₓ = Horizontal Component of Initial Velocity = ?

Therefore,

13.4 m = V₀ₓ(1.45 s)

V₀ₓ = 13.4 m/1.45 s

V₀ₓ = 9.2 m/s

Answer:

h = 51.75 m

nearest answer is:

53.4 m

Explanation:

First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance = 39 m

V₀ₓ = Horizontal Initial Velocity = 12 m/s

t = time = ?

Therefore,

39 m = (12 m/s)t

t = 39 m/12 m/s

t = 3.25 s

Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:

h = V₀y t + (1/2)gt²

where,

h = height of cannon = ?

V₀y = initial vertical velocity = 0 m/s

g = 9.8 m/s²

Therefore,

h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²

h = 51.75 m

Answer:

Yuh

Explanation:

Answer:

an object that is in motion wont go out of motion until there is another force pushing on it

Answer:

The modal verb must is used to express obligation and necessity. The phrase have to doesn't look like a modal verb, but it performs the same function.

Answer:

0.546 [tex]\hat k[/tex]

Explanation:

From the given information:

The force on a given current-carrying conductor is:

[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]

where the length usually in negative (x) direction can be computed as

[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]

[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]

[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]

where;

current I = 7.0 A

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]

F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]

F = 0.546 [tex]\hat k[/tex]

Answer:  Please see answer in explanation column.

Explanation:

Given that

v≈(331 + 0.60T)m/s

where Temperature, T =  14°C

v≈(331 + 0.60 x 14)m/s

v =331+ 8.4 = 339.4m/s

In our solvings, note that

f= frequency

 λ=wavelength

L = length

v= speed of sound

a) Length of the pipe is calculated using the fundamental frequency formulae that

f=v/2L

Length = v/ 2f

= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m

b) wavelength of the fundamental standing wave in the pipe

L = nλ/2,

λ = 2L/ n

λ( wavelength )= 2 x 0.3435/ 1

= 0.687m

c) frequency of the fundamental standing wave in the pipe

F = v/  λ

= 339.4m/s/0.687m=

494.03s^-1 = 494 Hz

d) the frequency in the traveling sound wave produced in the outside air.

This is the same as the frequency in the open organ pipe = 494Hz

e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m

f) To play D above middle c . the distance is given by

L =v/ 2 f

= 343/ 2 x 294

=0.583m

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

Answer:

The average power is  [tex]P = 195 .42 \ W[/tex]

Explanation:

From the question we are told that

   The  energy of the appliance is  [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]

    The time considered is   [tex]t = 1 \ day = 24 \ hours[/tex]

Generally the average power consumption is mathematically represented as

      [tex]P = \frac{E}{t}[/tex]

=>   [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]

=>   [tex]P = 195 .42 \ W[/tex]

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

Tension of the line = the weight + force from pulling up the fish

30 N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6 a

a = 1.7333333 m/s²

You are fishing and catch a fish with a mass of 6 kg. If the fishing line can withstand a maximum tension of 30 N, the maximum acceleration is  1.7333333 m/s².

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Answer:

a bowling ball because it has the most mass.

Answer:

Explanation:

pV = nrT

n = PV/RT

n = (100*10^3)(.01)/(300*0.082057)

n = 40.62 moles