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Aleksandr-060686 [28]
4 years ago
6

What is the mass of a crate if a force of 200 N causes t to accelerate at 8 ms? (Formula F=ma)

Physics
2 answers:
oee [108]4 years ago
6 0
Hey there! :D

Plug in what you know. 

F=ma

F= force m= mass a= acceleration 

200 N= m*8 ms

Divide both sides by 8. 

m= 25

The mass is 25. 

I hope this helps!
~kaikers 
Elden [556K]4 years ago
6 0
Here's what you know:
F = 200 N
a = 8 m/s^2

now, plug those numbers into the formula (f=ma) to get:
200 = m(8)

to solve for m, divide both sides by 8
Answer: 25

hope i helped :3
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A ball is thrown vertically upward with a speed of 1.86
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Answer:

t = 1.09 s.

Explanation:

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y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0

This quadratic equation can be solved using determinant.

\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s

Of course, we will choose the positive time.

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4 years ago
A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass
saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)         (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m

Furthermore, the angle α is:

\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°

Then, you solve the equation (1) for W and replace the values of all parameters:

W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J

The work done by the student is 609.97J

6 0
4 years ago
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