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luda_lava [24]
3 years ago
15

A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o

n the block. Calculate the force of friction between the block and the floor.​
Physics
1 answer:
Masja [62]3 years ago
3 0

Answers is  F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

               F=Ц N

where N=mg

putting values we get answer.

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A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig
SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

3 0
2 years ago
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
In establishing standing waves with a Slinky®, your wrist oscillates with f frequency of 4.0 Hz, and the length between any two
vladimir1956 [14]
Your answer would be <span>20.</span>
4 0
3 years ago
How to answer this question?
laila [671]

Answer:

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2 years ago
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