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Elis [28]
3 years ago
15

a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv

e? (b) what is its change in momentum?
Physics
1 answer:
Lina20 [59]3 years ago
7 0

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

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Answer:

C) three

Explanation:

Let gram of gold required be m . Let temperature change in both be Δ t .

heat absorbed = mass x specific heat x change in temperature

for copper

heat absorbed = 1 x .385 x Δt

for gold

heat absorbed = m x .129 x Δt

So

m x .129 x Δt = 1 x .385 x Δt

m = 2.98

= 3 g approximately .

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The density of mercury is 13.6 g/ml. what is its density in lbs/L
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3 years ago
Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550
Arisa [49]

Answer:

 q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2   (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

                            q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

                            w_poly = R*(T_f - T_i)/(1-n)

                            w_poly = 0.189*(350 - 400)/(1-1.2)

                            w_poly = 47.25 KJ/kg

-Hence,

                           q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

                           q_poly = 14.55 KJ/kg

4 0
3 years ago
Which is an example of potential en
SIZIF [17.4K]

a yoyo in someones hand is an example of potential energy

8 0
3 years ago
Read 2 more answers
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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