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Debora [2.8K]
3 years ago
12

If a change in speed “Δv” occurs and the mass starts at rest, would the change in speed also equal the final speed of the mass?

Physics
1 answer:
stealth61 [152]3 years ago
3 0

Yes, if the mass starts at rest, <u>the change in speed will be equal the final speed</u>, because:

Δv = Vf - Vo

How Vo (Initial velocity) is equal zero, we simplificate:

Δv = Vf

Then, the change of the speed, if the mass starts at rest, will be equal to final velocity.

Greetings.

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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
Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point
Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

B = \frac{\mu_oI}{2\pi R}

At the center, the magnetic field strength is twice

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Earths poles are on opposite sides of the earth
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bearhunter [10]

Answer:

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You may look at what group they are in
Group
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