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MAVERICK [17]
2 years ago
6

2. ____________________________ can cause a stationary object to start moving or a moving object to change its speed or directio

n or both.
Physics
1 answer:
Vikentia [17]2 years ago
7 0

As per Newton's law of inertia we can say every object will move in its state of motion either in the state of rest or will move with same constant velocity until some unbalanced external force will act upon it

so here we have to fill the space which says that stationary object can start or moving object will change its direction only in which case

so here the correct answer should be Unbalanced Force condition we will obtain such situation.

so we have

<u>Unbalance force</u> can cause a stationary object to start moving or a moving object to change its speed or direction or both.

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A scientist is designing a device that will mimic Earth's atmosphere by
mafiozo [28]

Answer:

1. Ultraviolet light (UV)

2. X-rays

3. Gamma-rays

Explanation:

Though there are different types of energy or electromagnetic waves with varying wavelengths, including the likes of Gamma X-rays, ultraviolet light, visible light, infrared radiation, and microwave radiation.

What is more certain is that the atmosphere blocked the high-energy waves from getting to the earth surface or biosphere such as Ultraviolet light (UV), X-rays and Gamma-rays

8 0
3 years ago
You are holding a finishing sander with your right hand. THe sander has a flywheel which spins counterclockwise as seen from beh
ryzh [129]

Answer:

c. turn downward

Explanation:

From the information given:

To find the tendency of the sander;

We need to apply the right-hand rule torque; whereby we consider the direction of the flywheel, the direction at which the torque is acting, and the movement of the sander toward the right.

Since the flywheel of the sander is in counterclockwise movement, hence the torque direction will be outward placing on the wall. However, provided that the movement of the sander is toward the right, then there exists an opposite force that turns downward which showcases the tendency in the sander is downward.

3 0
2 years ago
Amy has a mass of 30 kg, and she is riding a skateboard traveling 5 m/s. What is her momentum?
Flauer [41]
Hey there!

So we know that m*v=P.
And in this question m=30
v=5 m/s
P = 30*5 Kgm/s
P = 150 Kgm/s

So, your final answer is 150 Kg.m/s

Hope this helps! :)
5 0
3 years ago
Read 2 more answers
15 points for 2 questions
Elden [556K]
I believe the answer would be c because i think that you multiply the 2
8 0
2 years ago
Read 2 more answers
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
1 year ago
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