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myrzilka [38]
3 years ago
10

How does friction slow things down?

Physics
2 answers:
Lapatulllka [165]3 years ago
8 0
Friction slows things down because the tiny rough bumps on all surfaces hook on each other and slow down the object and produce heat
Ymorist [56]3 years ago
5 0
Friction slows things down because the force that is created through the interaction between two objects begins to decrease the acceleration and velocity of the object. Therefore, the object begins to slow down. 
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57:23
jok3333 [9.3K]

Answer:

X: Low potential energy

Y: High Potential energy

Z: Flow of electrons

Explanation: From the figure, it's obvious that Z is the flow of electrons, as shown by the arrow demonstrating the direction of the flow. Because of this, we can easily nullify choices B and C.

From the figure, we can notice that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.

8 0
3 years ago
If sulfur 34 undergoes alpha decay, what will I become?
ikadub [295]
It will decay into Silicon-30. Because alpha particles are 2 protons and 2 neutrons with an atomic mass of 4, you minus sulfur's atomic number by 2 and get silicon. And the atomic mass is 34 - 4 which equals 30.
3 0
3 years ago
A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00
Contact [7]

Answer:

2 m/s

Explanation:

m_1 = Mass of red truck = 1000 kg

m_2 = Mass of green truck= 3000 kg

u_1 = Initial Velocity of red truck = 6 m/s

u_2 = Initial Velocity of green truck

v = Velocity with which they move together = 0

For elastic collision

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s

Velocity of the green truck is 2 m/s

8 0
2 years ago
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is direc
Serga [27]

Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

8 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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