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almond37 [142]
2 years ago
14

A rocket sled is tested at "5 g" (5 times the acceleration due to gravity). If the sled starts from rest at position do= 0.00, h

ow long does it take to travel 441 meters? t = _____ s
Physics
2 answers:
stealth61 [152]2 years ago
8 0
D=at²
441m=(5*9.81m/s²)(t²)
t²=441/(5*9.81)
t≈√8.99
t≈3 sec
Tems11 [23]2 years ago
5 0

Answer:

It will take 4.24 seconds to travel 441 meters.

Explanation:

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case the velocity of body in vertical direction = 0 m/s, acceleration = 5g = 5 x 9.8 = 49 m/s^2, we need to calculate time when s = 441 meter.

Substituting

441=0\times t+\frac{1}{2} \times 49\times t^2\\ \\ t = 4.24 seconds

So it will take 4.24 seconds to travel 441 meters.

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Shkiper50 [21]
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Let's let +x be East and -x be West.

 -0.9 + 2.7 = 1.8
 
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What is meant by the following statement? "Acceleration is inversely proportional to mass." Acceleration decreases as mass decre
Vladimir79 [104]
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aev [14]

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6 0
2 years ago
A hopper jumps straight up to a height of 1.1 m. With what velocity did it leave the floor?
Amanda [17]

Answer:

4.64m/s

Explanation:

We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).

v = √2(9.8)(1.1)

v ≈ 4.64m/s

Best of Luck!

4 0
2 years ago
What is the coefficient of friction between the skates and the ice?
natta225 [31]

The coefficient of friction is 0.051

Explanation:

The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2 - u^2 = 2as

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

s=1.0\cdot 10^2 m = 100 m is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

F= ma = -\mu mg

where

-\mu mg is the force of friction

m is the mass of the skater

\mu is the coefficient of friction

a=-0.5 m/s^2 is the acceleration

g=9.8 m/s^2 is the acceleration of gravity

Solving for \mu, we find the coefficient of friction:

\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0
3 years ago
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