Ask question

Ask question

All categories

4 years ago

5

The policeman also watched a truck that was sitting at the red light. When the light turned green, the truck slowly started movi

ng, getting faster and faster until it was going a constant velocity.What do you think the motion graphs looked like?Explain why you think they look this way.

2 answers:

Juli2301 [7.4K]4 years ago

7 0

Answer:

I think the motion graphs would slope higher and higher as the truck goes faster and faster until it stops again.

Explanation:

DerKrebs [107]4 years ago

3 0

Answer:

a steady level climb until he gets to his constant speed

Explanation:

a steady level climb until he gets to his constant speed

You might be interested in

Please Help <br><br> Conservation of energy

Sonja [21]

Answer:

a principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

8 0

3 years ago

2. El sonido de una ballena es en especial de frecuencia baja, pero existe una especie de ballena la Whalien cuya frecuencia es

ikadub [295]

Answer:

The sound of a whale is especially low frequency, but there is a species of whale the Whalien whose frequency is 52 Hz, if the propagation speed of the wave is 1400m / s What will be its period in the water and the air? And what will be the wavelength in each medium? Remember that the propagation speed in air is 340m / s

Explanation:

From wave equation, the speed, wavelength and frequency is related using

V = fλ

Where

V is the speed

f is the frequency

And λ is the wavelength

So,

The frequency of the whale is

f = 52Hz

The speed in water is V_w = 1400m/s

The speed in air is V_a = 340m/s

We want to find the period in each medium, the period is related to the frequency and since the frequency is constant.

Then, period in equal in both medium

T = 1 / f

T_w = T_a = 1 / f

T = 1 / 52

T = 0.0192 seconds

We want to find the wavelength in each medium

For water,

V = fλ

V_w = f × λ_w

Then,

λ_w = V_w / f.

λ_w = 1400 / 52 = 26.92 m

The wavelength in water is 26.92m

Now, in air

V = fλ

V_a = f × λ_a

Then,

λ_a = V_a / f.

λ_a = 340 / 52 = 6.54 m

The wavelength in air is 6.54 m

In Spanish

De la ecuación de onda, la velocidad, la longitud de onda y la frecuencia se relacionan usando

V = fλ

Dónde

V es la velocidad

f es la frecuencia

Y λ es la longitud de onda

Entonces,

La frecuencia de la ballena es

f = 52Hz

La velocidad en el agua es V_w = 1400m / s

La velocidad en el aire es V_a = 340m / s

Queremos encontrar el período en cada medio, el período está relacionado con la frecuencia y dado que la frecuencia es constante.

Luego, período igual en ambos medios

T = 1 / f

T_w = T_a = 1 / f

T = 1/52

T = 0.0192 segundos

Queremos encontrar la longitud de onda en cada medio

Para agua,

V = fλ

V_w = f × λ_w

Entonces,

λ_w = V_w / f.

λ_w = 1400/52 = 26,92 m

La longitud de onda en el agua es de 26,92 m.

Ahora en el aire

V = fλ

V_a = f × λ_a

Entonces,

λ_a = V_a / f.

λ_a = 340/52 = 6,54 m

La longitud de onda en el aire es de 6.54 m.

8 0

3 years ago

At Which Location Would An Objects Weight Be the Greatest?

asambeis [7]

Answer:

As I know it is in Jupiter the objects weight more than 2x than on Earth.

Please give me 5* if this was helpful. :)

7 0

3 years ago

Read 2 more answers

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

A train traveling at 27.5 m/s accelerates to 42.4 m/s over 75.0 s. What is the displacement of the train in this time period

Sergio [31]

Answer:

2621.25 meters

Explanation:

First, write down what we are given.

Initial velocity = 27.5 m/s

Final velocity = 42.4 m/s

Time = 75 seconds

We need to look at the kinematic equations and determine which one will be best. In this case, we need an equation with distance. I am going to use $v_{f}^{2} = v_{i}^{2} +2ad$ , but you can also use the other equation, $x = v_{o}t+\frac{1}{2}at^{2}$

We need to find acceleration. To find it, we need to use the formula for acceleration: $a = \frac{v_{f}-v_{i}}{t}$ . Plugging in values, $a = \frac{42.4-27.5}{75} = .199\ m/s^{2}$

Next, plug in what we know into the kinematics equation and solve for distance. $42.4^{2} = 27.5^{2} + 2(.199)(d)\\d = 2621.25\ meters$

7 0

3 years ago