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emmasim [6.3K]
3 years ago
14

A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s

peed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)
a. gravity
b. force
c. force of friction
d. normal force
Physics
2 answers:
Monica [59]3 years ago
7 0
Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
              B. Force

If This Helps May I Have Brainliest?</span>
mel-nik [20]3 years ago
7 0
Hey there,

Your question states: <span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)

Your correct answer would be "force of friction" and also just plain "force". The reason why it would just be "force" would be because the student gave a little push to the block of dry ice. This would be the motion of force. Now, the fact that the block continued to move at a constant speed would be because it had a "force of friction" which means that there is a substance that would attract the block of dry ice. That is why the "force of friction" would be your answer. Now, GRAVITY  would be involved in this case, the block of ice is already on the ground, this gravity would NOT be your answer.

Hope this helps.

~ Nerdy Astute 

</span>
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collection of such objects we observe that they emit electromagnetic radiation of three different energies: 0.7 eV (infrared), 2
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It 45 because it has to be this and that and ur gay

4 0
3 years ago
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0
3 years ago
A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
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f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
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\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
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A psychopathologist records the number of criminal offenses among teenage drug users in a nationwide sample of 1,201 participant
lara [203]

Answer:

Standard deviation = 3

Explanation:

Given

N = 1201

SS = 10800

Required

Determine the standard deviation

First, we need to determine the variance;

Variance = \frac{SS}{N - 1}

This gives:

Variance = \frac{10800}{1201 - 1}

Variance = \frac{10800}{1200}

Variance = 9

Know that:

Variance = SD^2

Where SD represents standard deviation

This gives

9 = SD^2

Take square root

SD = \sqrt 9

SD = 3

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Brrunno [24]

Answer:

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Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.

The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.

Hope that helps.

Explanation:

5 0
3 years ago
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