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oksian1 [2.3K]
3 years ago
10

Which expression shows the amount of work a 6-kilogram mass is capable of performing if it is dropped from a height of 2 meters?

The acceleration of gravity is .
Physics
2 answers:
galina1969 [7]3 years ago
6 0
To answer this question you need to know that work = force x distance. You know the distance is 2m. Weight is a force and you can find that with the information given.
Weight = mass x gravity
In this case we're taking about acceleration of gravity which equals 9.8m/s/s
Weight = 6kg x 9.8m/s/s
W= 58.8 newtons or kg-m/s/s
So now that you know force and distance you can find work
Work= force x distance
Work = 58.8 newtons x 2 meters
Work = 117.6 n-m or joules
Hope this helps!
Ipatiy [6.2K]3 years ago
3 0
The acceleration of gravity is 9.8 m/s^2
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the force shown in figure 7-15 moves an object from x=0 to x=0.75 m. How much work is done by the force?
Phantasy [73]
Work is force multiplied by the distance the force moves the object
6 0
3 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

3 0
3 years ago
You throw a rock up into the air as hard as you can. It stays in the air a total of 6.0 s. What
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Answer:

<h2>5.9 x 10</h2>

Explanation:

6 0
3 years ago
A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to
nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to  the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

4 0
3 years ago
H<br>Ggghffyjfdudjhfhghggffghjjdxv
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Answer:

H

Ggghffyjfdudjhfhghggffghjjdxv

Explanation: The law of copying

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3 years ago
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