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bulgar [2K]
3 years ago
8

What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per

second (m/s) and assume acceleration from gravity is 10 m/s2.
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

speed = 40 m/s

Explanation:

Since the object is dropped, V0y = 0.

Vy = V0y - gt

= -(10 m/s^2)(4 s)

= -40 m/s

This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.

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Which of these warnings refers to a chemical property of the material?
Wewaii [24]

Answer:

Please show the warning's

Explanation:

6 0
4 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
During a softball game, a player running from second base to third base reaches a speed
andrey2020 [161]

Answer: I would say it would be 3.9

but i believe there is something missing shorty.

4 0
3 years ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
PLZ HELP ITS A TIMED TEST!!!!!!!!!!!!!!!!
sergij07 [2.7K]
Image #3 good luck!!!!!!!!!!!
7 0
3 years ago
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