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erastova [34]
2 years ago
10

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o

pposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
Physics
1 answer:
katen-ka-za [31]2 years ago
4 0

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

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Therefore, the acceleration of the ball is -2.5m/s²

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6 0
3 years ago
Moving the probe 1 cm towards the non-grounded electrode changes the value the potential from about 0.90 V to about 1.2 V. Expla
svp [43]

Answer:

-30 N/C

Explanation:

Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m

Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V

Since E = -ΔV/Δx

substituting the values of the variables into the equation, we have

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Since 1 V/m = 1 N/C.

E = -30 N/C

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The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

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M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

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6 0
3 years ago
an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei
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If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

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<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle  = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

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Please help <br>Newton second law of motion is F =ma <br>​
mamaluj [8]

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