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3 years ago

How did the coastal mountains form

1 answer:

4 0

Normally its when two tectonic plates hit each other and one plate subducts underneath the other this causes the other plate to be thrust upwards forming mountains

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A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

$v(t)=v_0 -gt$

where

$v_0=0$ is the initial vertical velocity (zero, since the motorcycle is not moving)

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

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

$v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s$

And since we know its mass, m=100 kg, we can find its momentum:

$p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s$

and the negative sign simply means downward direction.

8 0

3 years ago

Which statement about protons is true?

White raven [17]

GIVE vo0Mac0ov Brainliest ... He's smart

4 0

3 years ago

Read 2 more answers

We tend to feel less _____ when more people are around to help out. compassionate conformity responsible no answer is correct

never [62]

Responsible because we feel less stressed when others help out with our responsibilities.

5 0

4 years ago

The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s.

Klio2033 [76]

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

$v^2=u^2+2as$

Here final velocity, v = 26 m/s

a = acceleration due to gravity

$a=9.8m/s^2 \\$

displacement, s = 33 m

Substituting

$26^2=u^2+2\times 9.8 \times 33\\\\u^2=29.2\\\\u=5.40m/s \\$

Speed of water at the top of fall = 5.40 m/s

3 0

3 years ago

Please help on this one

Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

X Y

v₁ 32 cos50 m/s 32 sin50 m/s

v₂ 32 cos50 m/s ?

Δd ? 0

Δt ? ?

a 0 -9.8 m/s²

Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

Δdy = v₁yΔt + 0.5ay(Δt)²

0 = v₁yΔt + 0.5ay(Δt)²

0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

0 = v₁ + 0.5ayΔt

0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

0 = 24.513 m/s - 4.9m/s²Δt

-24.513m/s = -4.9m/s²Δt

-24.513m/s ÷ 4.9m/s² = Δt

5.00s = Δt

Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(5.00s) + 0.5(0)(5.00)²

Δdₓ = 32cos50m/s(5.00s)

Δdₓ = 102.846

Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.

6 0

3 years ago