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3 years ago

6

The airport has a new plane that can go from 0 miles per hour to 100 miles per hour in 10 seconds. What is being described about

the new plane?
a. acceleration
b. speed
c. velocity
d. magnitude

1 answer:

morpeh [17]3 years ago

7 0

Answer:

acceleration

Explanation: how fast something can get to a certain speed

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If a bike is traveling at 12 km/hr for 20 minutes how far will it travel

geniusboy [140]

1) 20 min = 1/3 h
2) distance = speed * time = 12 * 1/3 = 4 km

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3 years ago

A person stands between two speakers driven by the same source. Each speaker produces a tone with a frequency of 200 Hz on a day

MissTica

Answer:

The type of interference the person perceives is a Constructive Interference

Explanation:

Before we solve the problem we need to clarify some concepts:

There are two types of Interference, the Constructive and the Destructive interference

The Constructive Interference happens if the difference between the waves movement is equal to an even semi-wavelength number

The Destructive Interference is the opposite than the Constructive Interference, it happens if the difference between the waves movement is equal to an odd semi-wavelength number

Taking the above into account let´s proceed to solve the problem.

First, we will need to find how much the wavelength is (we will call it l), we know the frequency f=200Hz, we also know the speed of sound is v=330m/s. We will use the next equation:

$l=v/f$ then we get that $l=1.65m$

In order to find the semi-wavelength we must divide by 2 the previous result, in this case the semi-wavelength will be:

$l/2=0.825m$

Now, we find the difference between the waves movement, which is the difference between both distances, this difference will be called d and we will find it this way:

$d=4.95m-1.65m$ then d=3.3m

After this, we will do a relation between the distance and the semi-wavelength in order to find what type of interference the person perceives, this way:

$(d)/(l/2)$ then, we have a relation equals to 4. As it´s an even number, we determine the interference is a Constructive Interference

8 0

4 years ago

A car has a mass of 1300 kg and a velocity of 15 m/s. The car crashes into a wall and stops in two seconds. What is the stopping

Musya8 [376]

Answer:

B

Explanation:

Newton's second law:

F = ma

The acceleration is Δv/Δt, so:

F = m Δv/Δt

Given m = 1300 kg, Δv = 15 m/s, and Δt = 2 s:

F = (1300 kg) (15 m/s) / (2 s)

F = 9750 N

Answer B.

3 0

3 years ago

statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0

3 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

3 years ago