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3 years ago

What does the slope of a speed vs time graph means

Physics

1 answer:

Agata [3.3K]3 years ago

3 0

Answer:

speed = m/s on y axis

time = s on x axis

when you calculate the slope you do y2-y1/x2-x1 right?

m/s/s become $m/s^2$ , which is the unit of acceleration

On a speed vs time graph, if you see that the line is constant, then acceleration = 0 because there is no difference in speed, you are not acceleration or decelerating.

So in summary, it's acceleration

Hope that that answers your question

Don't hesitate to leave a comment if you are confused about something

Explanation:

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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

Imagine that you pushed a box, applying a force of 60 newtons, over a distance of 4 meters. how much would you have done? 15 jou

Ray Of Light [21]

The answer is 240 joules. Good luck!

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3 years ago

Read 2 more answers

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0

3 years ago

Question 26 Unsaved

vlabodo [156]

The density increases

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3 years ago

Which of the following astronomers was the first to propose elliptical orbits for the planets in our solar system?A.Brahe.Kepler

Anastasy [175]

B. Kepler, he made the laws for planetary motion

5 0

4 years ago

Read 2 more answers