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3 years ago

A 1.0 kg ball falls from rest a distance of 19.6 m.

Physics

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

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A moving rope (parallel to the slope) is used to pull skiers up the mountain. If the slope of the hill is 37" and friction is ne

Charra [1.4K]

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Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$

$W_y = 500 sin37$

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

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$F = 300 N$

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4 years ago

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago

While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?

xxTIMURxx [149]

Answer:

-2.83 m/s²

Explanation:

Initial velocity (u) = 34 m/s
Final velocity (v) = 17 m/s
Time taken (t) = 6 seconds

❖ Acceleration is defined as the rate of change in velocity with time.

→ a = (v - u)/t

v denotes final velocity
a denotes acceleration
u denotes initial velocity
t denotes time

→ a = (17 - 34)/6 m/s²

→ a = -17/6 m/s²

<h3>→ Acceleration = -2.83 m/s²</h3>

(Minus sign implies that the velocity is decreasing.)

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3 years ago

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3 years ago

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inessss [21]

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3 years ago