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Katyanochek1 [597]
3 years ago
5

A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the

normal force?
Physics
1 answer:
Natali [406]3 years ago
4 0
There are three forces acting on the book. 
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.
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Metallic bonds are responsible for many properties of metals, such as conductivity. Why is this possible? (1 point)
qaws [65]

Answer:

The bonds can shift because valence electrons are held loosely and more freely

Explanation:

Please give brainliest if you can,have a good day<3 :)

4 0
2 years ago
What happens when the thermal energy of a substance increases?
Sonja [21]

When thermal energy of a substance increases, it's entropy(randomness) & Kinetic energy increases.

For more appropriate answer, you should put the options 'cause there could be more than one answer for this question.

4 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
John and Caroline go out for a walk one day. This graph represents their distance from home.
Serga [27]

Answer:

Option C is the correct answer.

Explanation:

  We have velocity of a body = Change in position/ Time.

  Considering first portion of graph,

  Change in position = 30 - 0 = 30 m

  Time = 0.75 hours = 45 minutes = 2700 seconds

   Velocity = 30/2700 = 0.011 m/s

  Considering second portion of graph,

  Change in position = 30 - 30 = 0 m

  Time = 0.5 hours = 30 minutes = 1800 seconds

   Velocity = 0/1800 = 0 m/s

 Considering third portion of graph,

  Change in position = 0 - 30 = -30 m

  Time = 0.75 hours = 45 minutes = 2700 seconds

   Velocity = -30/2700 = -0.011 m/s

So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).

Option C is the correct answer.

3 0
3 years ago
Read 2 more answers
What measures the radiation given off by earth
gladu [14]
 i thinkits an instrument called seismograph. not sure
3 0
2 years ago
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