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myrzilka [38]
3 years ago
10

100 + 50 + BRAINLIST PLEASE HELP!!!

Physics
2 answers:
irina1246 [14]3 years ago
6 0

Answer:

Explanation:

1. What are the forces acting on the block when it is hanging freely from the spring scale? What is the net force on the block? What are the magnitudes of each of the forces acting on the block? Explain.

When a block is hanging freely, two forces are acting on it = tension force from the spring scale and gravity force on the block itself. The net force is zero as the block is not accelerating.  The magnitudes of tension and gravity force are the same but in opposite directions.

2. What are the forces that act on the block when it is placed on the ramp and is held in place by the spring scale? What is the net force acting on the block? Explain. (Assume that the ramps are frictionless surfaces.)

There are three forces acting on the block when it is placed on the ramp and is held in place by the spring scale: as in 1, there are tension and gravity but there is a third force - reaction force from the ramp surface on the block that is perpendicular to the surface. Again the block is not moving so the net force is zero.

3. What is the magnitude of normal force acting on the block when it is resting on the flat surface? How does the normal force change as the angle of the ramp increases? Explain. (Assume that the ramps are frictionless surfaces.)

On flat surface, the normal force is equal to the gravity force of the block i.e. its weight. On a vertical surface, the normal force is equal to zero. For the angle of ramp, θ, the normal force = weight * cos θ.

SOVA2 [1]3 years ago
3 0

Answer:

Explanation:

1. block is hanging freely - freely means there is no net force

from the spring scale - spring force

block - weight

magnitude of weight = magnitude of spring force as there is no net force

2. same thing - held in place means there is no net force

from the spring scale - spring force

block - weight

ramp - normal reaction

3. when its resting, there is no net force

magnitude of normal reaction force = magnitude of block weight

normal force reaction force decreases as angle of ramp increases from zero

reaction = weight x sin (ramp angle)

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Mashutka [201]

The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.

<h3>What is absolute pressure?</h3>

Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool,

mass of the two person = 150 Kg

weight of water level displaced exists equal to the weight of person.

\rho \mathrm{Vg}=2 \mathrm{mg} \\

V=\frac{2 m}{\rho} \\

V=\frac{2 \times 150}{1000} \\

\mathrm{~V}=0.3 \mathrm{~m}^3

Area of pool $=\frac{\pi}{4} d^2$

&=\frac{\pi}{4} \times 6^2 \\

&=28.27 \mathrm{~m}^2

Height of the water rise

h &=\frac{V}{A} \\

h &=\frac{0.3}{28.27} \\

& \mathrm{~h}=0.0106 \mathrm{~m}

  • Pressure increased

P = ρ g h

P = 1000 x 10 x 0.0106

P = 106.103 Pa

To learn more about absolute pressure refer to:

brainly.com/question/17200230

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1 year ago
State the law of refraction
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Answer:

The law of refraction states that the incident ray, the refracted ray, and the normal to the interface, all lie in the same plane.

Explanation:

8 0
3 years ago
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A baseball has mass 0.147 kg. If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55
Anit [1.1K]

Explanation:

We have,

Mass of a baseball is 0.147 kg

Initial velocity of the baseball is 44.5 m/s

The ball is moved in the opposite direction with a velocity of 55.5 m/s

It is required to find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Change in momentum,

\Delta p=mv-mu\\\\\Delta p=m(v-u)\\\\\Delta p=0.147\times ((-55.5)-44.5)\\\\\Delta p=-14.7\ kg-m/s\\\\|\Delta p|=14.7\ kg-m/s

Impulse = 14.7 kg-m/s

Therefore, the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.7 kg-m/s

4 0
3 years ago
The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i
balandron [24]

Answer:

The answer is

<h2>84.9 kPa</h2>

Explanation:

Using Boyle's law to find the final pressure

That's

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}

From the question

P1 = 115 kPa

V1 = 480 mL

V2 = 650 ml

So we have

P_2 =  \frac{115000 \times 480}{650}  = \frac{55200000}{650}  \\  = 84923.076923...

We have the final answer as

<h3>84.9 kPa</h3>

Hope this helps you

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A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during
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A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
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We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

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  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0
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