What units for time must be used in calculating power

While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it

diamong [38]

Answer: 6s

Explanation:

Vs=32m/s speed at beginning of slowing down

Vf=0m/s stop speed

a= -6 m/s² acceleration

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Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

7 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

3 years ago

Suppose you read in the newspaper that a new planet has been found. Its average speed in its orbit is 33 kilometers per second (

Harlamova29_29 [7]

Answer:

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

Explanation:

We can answer this question by using Kepler's second law of planetary motion, which states that:

"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"

This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).

In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

5 0

3 years ago

El ancho de la uña de su dedo meñique se acerca a una unidad metrica comun ¿cual?

BartSMP [9]

Answer:

I have no clue what's really going on I'm just here to get answer maybe I will just try to get an answer but I have no clue I'm sorry I am confused and dint really know what to do here.

3 0

3 years ago

Why do different substances have diffrent properties

Artist 52 [7]

Different elements require different levels of energy to make or break a bond

8 0

3 years ago

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