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ss7ja [257]
3 years ago
7

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before s

triking the floor, and a speed of 6.5 m/s just after bouncing, find the average force of the floor on the ball. Assume that the ball is in contact with the floor for 0.32 s, and that the mass of the ball is 0.42 kg
Physics
1 answer:
Ganezh [65]3 years ago
6 0

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

F = m*a

so find the change of momentum P

Pdown

P=m*v_1=0.42kg*3m/s

Pup

P=m*v_1=0.42kg*6.5m/s

dP = change in P

dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s

dT = 0.32 s

so

F = \frac{dP}{dt}=\frac{3.99Kg*m/s}{0.32s} =12.468 N

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A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
aleksandr82 [10.1K]

Answer:

A(t) = -340e^{-t/70} + 350

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

y = ce^{bt} + \frac{a}{b}

So A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

A(t) = -340e^{-t/70} + 350

4 0
3 years ago
3.<br> is the force on a dropped ball.<br> gravity<br> acceleration
Oduvanchick [21]

Answer:

Speeding up While Falling Down

Gravity is a force that pulls objects down toward the ground. When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion.

mark me as brainliest

Explanation:

5 0
3 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0
3 years ago
Hii! help asap. i’ll give brainliest thanks!
oksian1 [2.3K]

Answer:

a i believe

Explanation:

7 0
3 years ago
What effect does a tripling of the net force have upon the acceleration of the object?
dsp73
The formula of net Force is:F = mawhere m is the mass of the objecta is the acceleration of the object
thus, if we triple the net force applied to the object:
3F = maa = 3F / m
The acceleration is also tripled since the force is directly proportional to the acceleration.


4 0
3 years ago
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