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abruzzese [7]
3 years ago
6

You have landed on an unknown planet, Newtonia, and want to know what objects weigh there. When you push a certain tool, startin

g from rest, on a frictionless horizontal surface with a 12.0-N force, the tool moves 16.0 m in the first 2.00 s. You next observe that if you release this tool from rest at 10.0 m above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia, and what does it weigh on Earth?
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Explanation:

Given

Force applied F=12\ N

time taken t_1=2\ s

Displacement s=16\ m

using

s=ut+\frac{1}{2}at^2

u=initial velocity

s=displacement

t=time

16=0+\frac{1}{2}a(2)^2

a=8\ m/s^2

thus mass of body m=\frac{F}{a}=\frac{12}{8}=1.5\ kg

Next it is released from a height of h=10\ m

time taken to reach ground t_2=2.58\ s

using  s=ut+\frac{1}{2}at^2 in vertical direction

here acceleration is due to acceleration due to gravity(g') of the planet

as it at rest so u=0 here

10=0+\frac{1}{2}(g')(2.58)^2

g'=3.004\ m/s^2

Thus acceleration due to gravity on Newtonian is 3\ m/s^2

Weight of tool on Newtonia W'=mg'

W'=1.5\times 3.004=4.506\ N

Weight on Earth W=1.5\times 9.8=14.7\ N

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Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

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In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

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\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

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In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
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Answer:

D = -4/7 = - 0.57

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Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

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Put this value in equation (3), we get:

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Answer:

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We could use the following suvat equation:

s=vt-\frac{1}{2}gt^2

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t = 1.3 s

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