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3 years ago

A book is sitting on a desk. what best describes the normal force acting on the book

Physics

2 answers:

Margarita [4]3 years ago

7 0

the upward force the desk pushes on the book.

from the free body diagram of the book, it is clear that on the book, there are two forces acting.

1) normal force F by the desk in Upward direction.

2) force of gravity by earth in downward direction.

The Upward normal force by the desk on the book balances the force of gravity by earth in downward direction. that is why the book remains stationary on the desk.

blsea [12.9K]3 years ago

4 0

The forces acting upon the book are shown below. The force of gravity pulling downward and the force of the table pushing upwards on the book are of equal magnitude and opposite directions. These two forces balance each other.

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What would be the speed of the boxes when box b has fallen a distance of 0.50 m? The coefficient of kinetic friction between box

bija089 [108]

The given question is incomplete. The complete question is as follows.

Boxes A and B in the figure have masses of 13.4 kg. and 4.8 kg. , respectively. The two boxes are released from rest. What would be the speed of the boxes when box b has fallen a distance of 0.50 m? The coefficient of kinetic friction between box a and the surface it slides on is 0.20. Use conservation of energy.

Explanation:

Let us assume that box B is falling down, so equation of force for box B is as follows.

$m_{B}g - T = m_{B}a$

T = $m_{B}g - m_{B}a$ .......... (1)

Now, force of equation for box A is as follows.

$T - f_{k, A} = m_{A}a$

$T - \mu_{k}m_{A}g = ma$

T = $\mu_{k}m_{A}g + m_{A}a$ ........... (2)

We will equate both equation (1) and (2) as follows.

$m_{B}g - m_{B}a$ = $\mu_{k}m_{A}g + m_{A}a$

$m_{B}g - m_{B}a = \mu_{k}m_{A}g + m_{A}a$

$m_{B}g - \mu_{k}m_{A}g = a(m_{A} + m_{B})$

a = $\frac{4.8 kg - (0.20)(13.4 kg)}{13.4 kg + 4.8 kg} \times 9.8 m/s^{2}$

= 1.1415 $m/s^{2}$

Now, we will calculate the velocity of the box after falling through a distance of 0.50 m as follows.

$v_{B} = \sqrt{2aS}$

= $\sqrt{(2 \times 1.1415 \times 0.50)}$

= 1.07 m/s

Thus, we can conclude that speed of the given boxes is 1.07 m/s.

5 0

3 years ago

A transformer connected to a 120-V (rms) ac line is to supply 12,500 V (rms) for a neon sign. To reduce shock hazard, a fuse is

Tamiku [17]

Answer:

a) 104

b) 106

c) 884 mA

Explanation:

The ratio of the transformer is given by:

$N=\frac{V_{out}}{V_{in}}=\frac{12500V}{120V}=104$

We need to know the current in the primary in order to obtain the power applied.

$I_1=I_2*N=8.50mA*104=884mA\\\\P=I*V=884*10^{-3}A*120V=106W\\$

The current rating of the fuse is the current on the primary, 884mA as we calculated before in order to obtain the power.

4 0

3 years ago

Does temperature affect the amount of sugar that dissolved in tea? Why

Andrei [34K]

Answer:

Generally, a solute dissolves faster in a warmer solvent than it does in a cooler solvent because particles have more energy of movement. For example, if you add the same amount of sugar to a cup of hot tea and a cup of iced tea, the sugar will dissolve faster in the hot tea.

8 0

3 years ago

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, $v_{o} = 20.0 m/s$

Angle made by the ramp, $\theta = 22.0^{\circ}$

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH$

where

H = $dsin22^{\circ} = 5sin22^{\circ}$

g = $9.8 m/s^{2}$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ}}$ = 19.06 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m$

(b) now, considering the coefficient of friction bhetween ramp and the ball, $\mu = 0.150$ :

velocity can be given by the eqn-3 of motion:

$v^{2} = v_{o}^{2} - 2gH - \mu gd$

$v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5$

$v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}$ = 18.7 m/s

Now, maximum height attained is given by:

$h = \frac{(vsin\theta)^{2}}{2g}$

$h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m$

Height from the ground = $5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m$

6 0

3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0

3 years ago