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Art [367]
2 years ago
11

A tennis player hits a 1.45 kg tennis ball with a racket of mass 2.5 kg. If he hits the ball with a velocity of 7.5 m/s and then

stops, what impulse did he imply on the ball? What is the ball’s velocity?
Physics
1 answer:
Marta_Voda [28]2 years ago
3 0
I = mΔv
I = 2.5 x (7.5 - 0) = 18.75 kg.m/s

Now, considering that the ball´s initial velocity was zero, we have:

1.45 x 0 + 2.5 x 7.5 = 1.45 x V
1.45V = 18.75
V = 12.93 m/s

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Kuiper Belt objects are composed of which substances? Select all that apply.
levacccp [35]
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron

8 0
3 years ago
Use V=Voe^-t/RC and T=RC to derive T1/2=Tln2
Ksju [112]
Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.

3 0
3 years ago
How do you calculate the mass of an object accelerating at22.35m/s2 with a force of 120N
Arte-miy333 [17]

Answer:

The mass of object is calculated as 5.36 kg

Explanation:

The known terms to find the mass are:

           acceleration of object (a) = 22.35 m/s^{2}

                        Force exerted (F) = 120N

                        mass of an object (m) = ?

From Newton's second law of motion;

                                   F = ma

                           or, 120 = m × 22.35

                          or, m= \frac{120}{22.35} kg

                           ∴ m = 5.36 kg

3 0
2 years ago
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
levacccp [35]

Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\

So, the minimum speed must the car must be 13.13 m/s.

5 0
2 years ago
Consult Multiple-Concept Example 5 to review the concepts on which this problem depends. A light bulb emits light uniformly in a
den301095 [7]

Answer:

a. 0.332 W/m² b. 11.2 V/m c. 15.8 V/m

Explanation:

(a) the average intensity of the light,

Intensity, I = P/A where P = average power = 150.0 W and A = area through which the power emits = 4πr² where r = distance from bulb = 6 m.

So, I = P/A = P/4πr²

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

I =  P/4πr²

I =  150.0 W/4π(6 m)²

I =  150.0 W/4π(36 m²)

I =  150.0 W/452.39 m²

I =  0.332 W/m²

(b) the rms value of the electric field,

Since Intensity, I = E²/cμ₀ where E = rms value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m.

Making E subject of the formula, we have

E² = Icμ₀

E = √(Icμ₀)

Since I = 0.332 W/m², substituting the other terms into the equation, we have

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(0.332 W/m² × 3 × 10⁸ m/s × 4π × 10⁻⁷ H/m.)

E = √(12.5 × 10)

E = √125 V/m

E = 11.18 V/m

E ≅ 11.2 V/m

(c) the peak value of the electric field.

The peak value of electric field, E' is gotten from E = E'/√2 where E = rms value of electric field.

So, E' = √2E

= √2 × 11.2 V/m

= 15.81 V/m

≅ 15.8 V/m

6 0
3 years ago
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