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11111nata11111 [884]

1 year ago

13

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

sure is 1.0 × 10^5 Pa (1.0 × 10⁵ Pa). Calculate

1 answer:

timurjin [86]1 year ago

5 0

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

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Al llegar a detenerse, un automóvil deja marcas de derrape de 92m de largo sobre una autopista. Si se supone una desaceleración

Bezzdna [24]

Answer:

Explanation: no se español

8 0

2 years ago

In a game of egg-toss, you and a partner are throwing an egg back and forth trying not to break it. Given your knowledge of mome

lutik1710 [3]

F=dP/dt. So you want the momentum to change as slowly as possible in time to minimize the force. So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop. If you hold your hand steady when you catch it the force due to the impact could break it.

3 0

1 year ago

Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitationa

ElenaW [278]

Answer:

1.07 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1 = 200kg

Mass 2 = 500kg

Distance of separation = 25m

Unknown:

Gravitational attraction between the two bodies = ?

Solution:

To solve this problem, we use the equation of the universal gravitation;

F = $\frac{G mass 1 x mass 2}{r^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹Nm²kg⁻²

r is the distance

Now insert the parameters and solve;

F = $\frac{6.67 x 10^{-11} x 200 x 500 }{25^{2} }$ = 1.07 x 10⁻⁸N

8 0

1 year ago

A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

6 0

2 years ago

A flying mosquito hits the windshield of a moving car and gets smashed, but the car is intact. Which of the following statements

zhannawk [14.2K]

Answer:

Action and reaction are the same

Explanation:

Let us carefully analyze the situation in the context of Newton's third law, when the car hits the mosquito exerts a force that acts on the mosquito, ACTION and the mosquito responds to the force that is being applied with a force of equal magnitude and direction or opposite applied to the car (Reaction).

Consequently, the correct answer is: The force of the car on the mosquito is of the same magnitude as that of the mosquito on the car

6 0

1 year ago