A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area

of contact is 2000 cm2.

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

Answer:

Ke- = Ke+ = 0.294MeV

Explanation:

To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:

2[tex]E_p=2E_o+K_{e^-}+K_{e^+}[/tex]   (1)

Ep: photon energy = 0.804MeV

Eo: rest energy of one electron (and positron) = 0.51MeV

Ke-: kinetic energy of electron

Ke+: kinetic energy of positron

You replace the values of Ep and Eo in the equation (1):

[tex]K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV[/tex]

Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:

[tex]K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV[/tex]

Answer:

14.68m/s

Explanation:

As per the question, the data provided is as follows

Mass = M = 0.148 kg

Height = h = 11 m

Initial velocity = U = 0 m/s

Final velocity = V

Gravitational force = F

Mass = M

Based on the above information, the speed that hit to the ground is

As we know that

Work to be done = Change in kinetic energy

[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]

[tex]V^2 - U^2 = 2gh[/tex]

[tex]V^2 - 0 = 2gh[/tex]

[tex]V = \sqrt{2 g h}[/tex]

[tex]= \sqrt{2\times9.8\times11}[/tex]

= 14.68m/s

Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

Answer:

[tex]T_{1}[/tex] = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about [tex]T_{1}[/tex] we have:

(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0

⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)

[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

[tex]T_{2}[/tex]= 59.535 ÷ 2.4

[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23

[tex]T_{1}[/tex] = 14.88 N

Answer:

Explanation:

  a )

from lens makers formula

[tex]\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})[/tex]

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

[tex]\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})[/tex]

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]

[tex]\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}[/tex]

[tex]\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}[/tex]

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

Answer:

a. SO2

Explanation:

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters )  is the airplane above the ground 13 seconds after takeoff?

Answer:

The height  is [tex]H = 318.5 \ m[/tex]

Explanation:

From the question we are told that

   The speed at which the plane takes off is  [tex]u = 49 \ m/s[/tex]

      The angle at which it takes off is  [tex]\theta = 30 ^o[/tex]

        The time taken is [tex]t = 13 s[/tex]

The vertical distance traveled is  mathematically represented as

          [tex]H = u sin \theta t[/tex]

Substituting values  

         [tex]H = (49) * sin (30) *13[/tex]

        [tex]H = 318.5 \ m[/tex]

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

Answer:

Explanation:

Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...

Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...

Answer:

0.173 N.

Explanation:

We will calculate the mass and then use the following calculations on the surface of planet X that is :

                           [tex]W=mg[/tex]

We would use the following equation to get the value of g for planet X that is :

                   [tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]

Then, put the values in the above equation.

                          [tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]

                           [tex]\bf\mathit{g=3.80\;m/s^2}[/tex]

Now, we will measure the ball weight on planet X's surface:

                          [tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]

Then, we have to put the value in the above equation.

                        [tex]W=0.1\times 1.73=0.173\:N[/tex]

Speed = (distance) / (time)

Speed = (

Velocity = speed, and its direction

The velocity of the plane is 10.2 miles per second East.

(about 48 times the speed of sound)

Answer:

c. 5.50 km

Explanation:

8 min * 1h/(60 min) = 8/60 = 2/15 h

15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) =  1/240 h

8 min 15 sec = (2/15+1/240)h

40 km/h *(2/15 +1/240)h =5.50 km

Answer: 5.50 km

Explanation:

Answer:

1008.57kg/m3

Explanation:

Now the mass of fresh water is 250×1000 /1000000 = 0.25kg

Now the mass of salt water is

100×1030 /1000000 = 0.103kg

Note Density = mass / volume

Mass = volume × density

Note that converting from cm3 to m3 we divide by 1000000

Total mass = 0.25kg +0.103kg= 0.353kg.

Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3

Hence the density of the mixture= total mass / total volume

0.353kg/35 × 10^{-5}m3=1008.57kg/m3

Answer:

C = 74.53°

Explanation:

Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as

A.B = |A||B|cosC; let C be the angle between the vectors

12 = 5×9 cos C

Hence cos C = 12/45

C = cos^-1(12/45)

C = 74.53°

Answer:

The curve should be banked at an angle of 13 degrees.

Explanation:

We have,

Radius of a highway curve is 274 m

Speed of car on this curve is 25 m/s

Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.

[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]

g = 10 m/s²

Plugging all the values in above formula,

[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]

So, the curve should be banked at an angle of 13 degrees.

Answer:

Loudness: decreases

Amplitude: decreases

Pitch: stays the same

Frequency: stays the same

Explanation:

1.

An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.

2.

Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.

3 and 4.

The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.

Hope this helps!

Answer:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

[tex]\vec{F}=m\vec{a}[/tex]

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]

Both x and y component must be equal in the previous equality, then you have:

[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]

Hence, the vector F3 is:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Answer:

a = 5.34 m/s²

T = 37.86 N

Explanation:

This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:

a = g(m₂ - m₁)/(m₂ + m₁)

where,

a = acceleration of both masses = ?

g = 9.8 m/s²

m₂ = heavier mass = 8.5 kg

m₁ = lighter mass = 2.5 kg

Therefore,

a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)

a = (9.8 m/s²)(6 kg)/(11 kg)

a = 5.34 m/s²

The formula for tension in cable is derived to be:

T = 2m₁m₂g/(m₁ + m₂)

T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)

T = 37.86 N

Answer:

It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4

Answer:

The distance from the cornea vertex to the retina is 2.36 cm

Explanation:

The question is incomplete.

The complete question is as follows;

A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.

If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?

Solution.

We use image-object reaction to calculate the distance from the cornea vertex to the retina.

Mathematically;

n1/s + n2/s’ = n2-n1/R

From the question, we identify the following;

n1 ; Refractive index of air = 1

n2 ; Refractive index of lens = 1.4

S ; Object Distance = 36 cm

S’ = ?

R ; Radius of curvature of the cornea = 0.65

Substituting these values into the equation above;

1/36 + 1.4/S’ = (1.4-1)/0.65

{S’+ 36(1.4)}/36S’ = 0.4/0.65

{S’ + 50.4}/36S’ = 0.62

S’ + 50.4 = 22.32S’

50.4 = 22.32S’ -S’

21.32S’ = 50.4

S’ = 50.4/21.32

S’ = 2.36 cm

Answer: 36, 200 feet deep according to information on google

Explanation:

A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go

Answer:

173.75 km/hr in the NW direction.

Explanation:

Velocity is the time rate of change in displacement of a body. Mathematically:

v = d / t

where d = displacement

t = time

Therefore, the velocity of the car is:

v = 556 / 3.2 = 173.75 km/hr

The velocity of the car is 173.75 km/hr in the NW direction.

The velocity of a car will be "173.75 km/hr".

The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.

Displacement, d = 556 km

Time, t = 3.2 hours

We know the relation,

→ Velocity = [tex]\frac{Displacement}{Time}[/tex]

or,

→ V = [tex]\frac{d}{t}[/tex]

By substituting the values, we get

      = [tex]\frac{556}{3.2}[/tex]

      = [tex]173.75[/tex] km/hr

Thus the response above is right.

Find out more information about velocity here:

https://brainly.com/question/6504879

Answer:

490 N

Explanation:

is the correct answer

If the up and down vectors are the same length. The right vector is longer than the left vector, then  the net force acting on Hector and the toboggan would be 490 Newtons.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

As given in the problem John pushes Hector on a plastic toboggan .The free-body diagram is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 490 N. The second vector is pointing right, labeled F Subscript t Baseline = 735 N. The third vector is pointing upward, labeled F Subscript N Baseline = 490 N. The fourth vector is pointing left, labeled F Subscript f Baseline = negative 245 N.

The net force acting on the vertical direction = 490-490

                                                                           =0

The net force acting on the horizontal direction = 735 -245

                                                                                =490 Newtons

Thus, the net force acting on Hector and the toboggan would be 490 Newtons.

Learn more about Newton's second law from here, refer to the link ;

brainly.com/question/13447525

#SPJ5

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  [tex]\mu mg = \frac{mv^2}{r}[/tex]

    where [tex]\mu[/tex] is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation: