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3 years ago

Please help

Physics

1 answer:

stira [4]3 years ago

5 0

It’s D. The pitch would be lower. I’m 100% sure. I just took the test. I hope this helps you!! <3

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An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

Sergeeva-Olga [200]

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = 735 J/kg/C (3 sf)

4 0

3 years ago

You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen

gayaneshka [121]

Answer:

option B

Explanation:

When you rub a clear plastic pen with the wool the plastic pen gets charges this phenomenon is known as frictional charging.

Due to rubbing, the pen gets negatively charged.

We know, opposite charge attract each other and the same charge repel each other.

So, when the pen is negatively charged the tape might be positively charged or the tape might be uncharged.

Hence, the correct answer is option B

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3 years ago

The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b

Serggg [28]

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

Let's plug in our given values and see what we get!

340

440

−

0.773

3 0

2 years ago

A 10 kg cart is rolling across a parking lot with a velocity of .5 m/s. What is its momentum?

-Dominant- [34]

Hello!

$\large\boxed{p = 5 \text{ } kgm/s}$

Use the equation for momentum:

$p = mv$

Plug in the given mass and velocity into the equation:

$p = 10 * 0.5$

$p = 5 kgm/ s$

4 0

2 years ago

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

3 0

3 years ago