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vesna_86 [32]
3 years ago
12

Please help! Question is attached.​

Physics
1 answer:
il63 [147K]3 years ago
6 0

Answer:

a. 6

b. 6 m/s²

c. 300 m to the right

d. 30 secs

Explanation:

slope = rise /run

60-0/10-0

= 6

b. slope = acceleration = 6 m/s²

c. d=ut+1/2at²

t=10 (segment A last for 10 secs)

u - initial velocity = 0

so d = 0(10)+1/2*6*10²

=300 m

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Answer:

An object in a motion when it is continuously changing its position based on a reference point and observed by a person or a device.

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3 years ago
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OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

W = F.d.\cos\theta

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

W = F.d

Given:

F = 75 N

m = 31.8 kg

Final velocity v_f = 15.3 \frac{m}{s}

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\

Now we can determine the displacement from the following formula:

d = \frac{1}{2}a^2+v_0t+d_0

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

d = \frac{1}{2}at^2\\

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m

and the corresponding work:

W = F\cdot d=75 N\cdot 48.94 m =3670.5J


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