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2 years ago

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An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately af

ter the explosion a fragment of mass 10 kg has a velocity of 120 m/s straight downward. How high above the point of the explosion does the larger fragment rise?

1 answer:

olga_2 [115]2 years ago

7 0

Answer:

9654.34 m

Explanation:

from conservation of momentum

$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$

And from Conservation of Energy

$\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)$

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Which of the following best demonstrates the effect of static friction? A. A person dropping a ball to the ground. B. A person p

Dominik [7]

Answer:

C.

Explanation:

A person pushing a couch will face resistive force of friction . When resistive force is greater then his force of effort , couch will not move. This force is static friction because the couch is stationary. When the force of effort is increased , magnitude of static friction also increases keeping the couch stationary. The ability of the static friction to increase its magnitude is limited by a maximum value beyond which the couch starts moving. The static friction is then converted into kinetic friction.

In rest of the three cases object is already moving so kinetic friction is in action.

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2 years ago

A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.

kenny6666 [7]

Answer:

a) $|\Delta E|=4.58\: J$

b) $F=61.90\: N$

Explanation:

a)

We can use conservation of energy between these heights.

$\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})$

$\Delta E=0.608*9.81(0.6026-1.37)$

Therefore, the lost energy is:

$|\Delta E|=4.58\: J$

b)

The force acting along the distance create a work, these work is equal to the potential energy.

$W=\Delta E$

$F*d=mgh$

Let's solve it for F.

$F=\frac{mgh}{d}$

$F=\frac{0.608*9.81*1.37}{0.132}$

Therefore, the force is:

$F=61.90\: N$

I hope is helps you!

6 0

2 years ago

A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to

schepotkina [342]

Answer:

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E = 3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J

4 0

3 years ago

What is the acceleration of an object with a mass of

aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

<u>Δ Being Δ</u>

F = Force = 183 N

m = Mass = 367 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=183 \ N / 367\ kg}$

⇒ Resolving

$\boxed{ a = 0.49 \ m/s^{2}}$

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

3 0

3 years ago

A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh

Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

$\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .....(I)

Where, $\delta t'$ = proper time

$\delta t$ = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

$\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}$

For the second interval,

Put the value of $\Delta t'$ in the equation (I)

$\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value in the equation

$\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}$

$\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}$

$\Delta t_{2}=-5.77\times10^{8}\ s$

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0

3 years ago