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fenix001 [56]
3 years ago
12

A ball with a mass of 5 kg is acted upon by a force of 20 N. If the ball begins moving from a velocity of 0 m/s and accelerates

for 2 seconds, what is its change in velocity?
A. 0 m/s B. 2 m/s C. 4 m/s. D. 8 m/s
Physics
1 answer:
Strike441 [17]3 years ago
8 0
The answer is d

F=ma. so by substituting we find that a=4

and a=vi-vf/t

by substiting we find vf=8

so the difference is vf-vi = 8m/s
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Charlie drove around the block at constant velocity, is it true or false?​
Paha777 [63]

Answer:

FALSE I THINK

Explanation:

7 0
3 years ago
A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t
Mandarinka [93]

Answer:

The mmf required is 1.125×10^{-3} A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

F_{M} = Hl\\

where F_{M} is the Magnetomotive force (mmf)

H is the Magnetic field strength

l is the magnetic length

The magnetic permeability μ is given by

μ = B / H

Where B is the Magnetic flux density

and H is the Magnetic field strength

From the question,

B = 1.2Wb/m^2

μ = 1600m

From μ = B / H

∴H = B/μ

H = 1.2 / 1600\\

H = 7.5 × 10^{-4}A/m

Now, for the Magnetomotive force (mmf)

F_{M} = Hl\\

From the question

l = 1.5 m

∴ F_{M} = 7.5×10^{-4} × 1.5

F_{M} = 1.125×10^{-3} A

Hence, The mmf required is 1.125×10^{-3} A

3 0
3 years ago
How much force is required to accelerate a 4 kg bowling ball from 0 m/s to 2 m/s in 1 second? what amount of energy does the bow
Vladimir [108]
Well, F = ma

and a= change in vel / change in time
        = (2-0 ) / 1  = 2 m^2/s

so, F= 4* 2 = 8 N

and W = F.S  = 8 * 2 = 16 J

hope it helped
5 0
3 years ago
QUICKLY I AM IN THE LESSON NOWWW
vladimir2022 [97]

Answer:

I don't really know much about the whole stars thing, but I've written a lot of paragraphs in science

Explanation:

Remember:

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Include:

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7 0
3 years ago
A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top
Anton [14]

Answer:

y_{max} = 0.829\,m

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

v = 2.913\,\frac{m}{s}

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}

After some algebraic handling, a second-order polynomial is formed:

12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

The roots of the polynomial are, respectively:

\Delta s_{1} \approx 0.128\,m

\Delta s_{2} \approx -0.099\,m

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

\Delta s \approx 0.128\,m

The maximum height that the block reaches after rebound is:

(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}

y_{max} = 0.829\,m

4 0
3 years ago
Read 2 more answers
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