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Marat540 [252]
3 years ago
11

A car initially traveling at 32.4 m/s undergoes a constant negative acceleration of magnitude 1.70 m/s2 after its brakes are app

lied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?
Physics
1 answer:
8090 [49]3 years ago
8 0
<h2>The car tire makes 149 revolutions before coming to stop.</h2>

Explanation:

a) We have equation of motion v² = u² + 2as

Initial velocity, u = 32.4 m/s  

Acceleration, a = -1.70 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 32.4² + 2 x -1.70 x s

s = 308.75 m  

Distance traveled before stopping is 308.75 m

Radius of tire = 0.330 m

Circumference of tire = 2πx 0.33 = 2.07 m

1 revolution = 2.07 m

\texttt{Number of revolutions = }\frac{308.75}{2.07}=148.91

The car tire makes 149 revolutions before coming to stop.

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A person's _____________ will change if they move from the Earth to the moon.
Diano4ka-milaya [45]

1. A person's weight will change if they move from the Earth to the moon.

In fact, the weight of a person is given by:

W=mg

where m is the mass of the person and g is the gravitational acceleration. The mass of the person, m, is the same on the Earth and on the moon, but the value of g is different on the Moon (about 1/6 of the Earth's value), so the weight also changes.


2. An astronaut is launched into space. The mass of the astronaut did not change. This is measured in Kg.

The mass of an object (or of a person, as in this case) is an intrinsec property of the object, that depends on the amount of matter inside the object: therefore, this quantity does not depend on the location of the object, so it is the same on the Earth, on the Moon and in space.


3. What is the weight of a ring tailed lemur that has a mass of 10 kg? -98 N

The weight of the lemur is given by:

W=mg

where m=10 kg is the lemur's mass and g=-9.8 m/s^2 is the gravitational acceleration. Using these numbers, we find

W=(10 kg)(-9.8 m/s^2)=-98 N

and the negative sign simply means that the direction of the weight is downward.


4. What is the mass of the lemur from the previous question if it was on the International Space Station? 10 kg

As we said in question 1), the mass of an object does not depend on the location, so the mass of the lemur is still 10 kg, as in the previous exercise.


5. A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws? Third's Newton Law

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the fuel and the rocket. The fuel is pushed backward by the rocket, so the fuel exerts an equal and opposite force on the rocket, which then moves forward.


6. When a cannon is fired, the projectile moves forward. According to Newton's 3rd law, the cannon will want to travel backward.

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the cannon and the projectile.The projectile is pushed forward by the cannon, so the projectile exerts an equal and opposite force on the cannon, which moves backward.


7. An object has a weight of 21,532 N on Earth. What is the mass of the object? 2,197 kg

The weight of the object is given by: W=mg

If we re-arrange the formula and we use W=21,532 N, we can find the mass of the object:

m=\frac{W}{g}=\frac{21,532 N}{9.8 m/s^2}=2,197 kg


8. What is the mass of the object from the previous question if we put it on the moon? The force of gravity on the moon is 1.62 m/s2.  2,197 kg

As we said in question 4), the mass of an object does not change if we move it to another location, so its mass is still 2,197 kg.


9. How much force is exerted if a 250 kg object has an acceleration of 750 m/s2 ? 187,500 N

The force exerted on the object is given by Newton's second law:

F=ma

where F is the force, m=250 kg is the mass and a=750 m/s^2 is the acceleration. By using these numbers, we find

F=(250 kg)(750 m/s^2)=187,500 N


10. A resting soccer ball moving after it is kicked is an example of which of Newton's laws? Newton's second law

Newton's second law states that when an object is acted upon unbalanced force, the object has an acceleration, given by the law

F=ma

So, in this case, the ball is kicked and so an unbalanced force is applied to it, and for this reason the ball has an acceleration (in fact, it starts from rest, but then its velocity increases since it starts moving).

5 0
3 years ago
A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi
Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

where

ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

ρ = \frac{m}{V}

V = ρ/m

V = \frac{5.25.10^{5}}{10^{3}}

V = 525 m³

To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

r = \sqrt[3]{\frac{525.3}{4\pi } }

r = 5.005 m

diameter = 2*r = 10.01m

(a) Height for House A:

h = 15 + 10.01

h = 25.01

P = hρg

P = 25.01.10³.9.8

P = 245.10³ Pa or 245kPa

(b) h = 25 - 7.3

h = 17.71

P = hρg

P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

4 0
2 years ago
Momentum,
Serggg [28]

Answer:

THE WALL MOVES AWAY FROM THE BALL

Explanation:

NEWTON'S THIRD LAW STATES THAT THERE IS A OPPOSITE REACTION

5 0
2 years ago
The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea
djverab [1.8K]

Answer:

Acceleration, a=9.39\ m/s^2

Explanation:

Given that,

Mass of the planet Krypton, m=8.8\times 10^{23}\ kg

Radius of the planet Krypton, r=2.5\times 10^{6}\ m

Value of gravitational constant, G=6.6726\times 10^{-11}\ Nm^2/kg^2

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}

a=9.39\ m/s^2

So, the value of acceleration of an object in free fall near the surface of Krypton is 9.39\ m/s^2

5 0
3 years ago
If the universe is infinite, then how come there are things disappearing from the horizon? I was at a point in this video (TRUE
True [87]

Answer:

The big rip theory

Explanation:

      I believe what you are referring to is the big rip theory, in which the universe expands faster than the speed of light Kurzgesagt refers to it as a "horizon" but in reality it's a little more complicated than that. Eventually the expansion of the universe will accelerate far beyond the speed of light creating space between molecules until eventually all matter is fleeting and the entire universe is an endlessly vast cosmic void with not but the occasion molecule left from a time when things weren't so lonely.

3 0
2 years ago
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