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navik [9.2K]
3 years ago
12

The climber dropped her compass at the end of her 240-meter climb. How long did it take to strike bottom?

Physics
2 answers:
Maru [420]3 years ago
6 0

Answer:

7 seconds

Explanation:

Let's first write down what we have so that we can choose the correct equation of linear motion

initial velocity (u) = 0 m/s

distance (x) = h = 240 m

acceleration (a) = g = 9.81 m/s²

s = ut + \frac{at^{2} }{2} \\\\240 = (0)(t) + \frac{9.81t^{2} }{2}\\ \\9.81t^{2} = 480\\\\t = 6.99s\\\\t = 7 s

ANEK [815]3 years ago
4 0

240 = 0+1/2 (-9.8t

240 = -4.9t

<span>240/-4.9 = t</span><span>
</span>49.0 = t

t= 7.0s

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Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much
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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
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Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

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Horizontal force = Fcosθ = 250 cos35° = 204.79N

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Substituting

        204.79 = m x 1.4  

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3 0
3 years ago
Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan
Eddi Din [679]

To increase the centripetal acceleration to 2.00 m/s^2, you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

a=0.50 m/s^2

We want to increase it by a factor of 4, i.e. to

a'=2.00 m/s^2

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

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This way, the new acceleration is

a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

r'=\frac{1}{4}r

So the new acceleration is

a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

brainly.com/question/2562955

brainly.com/question/6372960

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5 0
3 years ago
**Fill in the blanks**
Nataly_w [17]

Answer:

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4 0
3 years ago
Read 2 more answers
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