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kramer
3 years ago
11

How long does it take for an 8 kg pumpkin to hit the ground if dropped from a height of 55 m.

Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

t = 3.35 s

Explanation:

It is given that,

Mass of a pumpkin, m = 8 kg

It is dropped from a height of 55 m

We need to find the time taken by it to hit the ground.

Initial velocity of the pumpkin, u = 0

Using second equation of motion to find it as follows :

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2(55)}{9.8}} \\\\t=3.35\ s

So, it will take 3.35 seconds to hit the ground.

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The Jamaican bobsled team was moving at a velocity of 50 m/s, then they hit the brakes on their sled to decelerate at a uniform
atroni [7]

Answer:

The time it took the bobsled to come to rest is 10 s.

Explanation:

Given;

initial velocity of the bobsled, u = 50 m/s

deceleration of the bobsled, a = - 5 m/s²

distance traveled, s = 250 m

Apply the following kinematic equation to determine the time of motion of the bobsled;

s = ut + ¹/₂at²

250 = 50t + ¹/₂(-5)t²

250 = 50t - ⁵/₂t²

500 = 100t - 5t²

100 = 20t -t²

t² - 20t + 100 = 0

t² -10t - 10t + 100 = 0

t (t - 10) - 10(t - 10) = 0

(t - 10)(t - 10) = 0

t = 10 s

Therefore, the time it took the bobsled to come to rest is 10 s.

3 0
3 years ago
PLEASE HELP!!!!I IM GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (30pts)
maksim [4K]

<em>Kinetic Energy</em>

=><em><u>It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.</u></em>

<em>Potential</em><em> </em><em>Energy</em><em> </em>

<u><em>=</em><em>></em><em>potential energy is the energy held by </em></u><em><u>an</u></em>

<em><u> object because of its position relative to </u></em><em><u>other</u></em>

<em><u> objects, stresses within itself, its </u></em><em><u>electric</u></em>

<em><u> charge, or other factors.</u></em>

<h2>Difference:</h2>

=>Potential energy is a <u>stored</u> energy on the other hand kinetic energy is the energy of an object or a system's particle in <em><u>Motion</u></em>.

6 0
3 years ago
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

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3 0
3 years ago
Read 2 more answers
A current of 1.4 A flows in a conductor for 7.0 s. How much charge passes a given point in the conductor during this time?
bija089 [108]

Answer:

The charge passes a given point in the conductor during this time is 9.8 C.

Explanation:

Given that,

Current = 1.4 A

Time = 7.0 sec

We need to calculate the charge during this time

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Charge is the product of current and time.

In mathematically form,

Q = i\times t

Where, i = cirrent

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Put the value into the formula

Q =1.4\times7.0

Q=9.8\ C

Hence, The charge passes a given point in the conductor during this time is 9.8 C.

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Answer:

True

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This means that induced current opposes the very cause that produces it.

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