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Bingel [31]
3 years ago
8

A car starts from rest and after 10 seconds is traveling at 20 m/s. Assuming that it continues to accelerate at the same rate it

will take another ________ seconds to reach 40 m/s.
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.

Answer:

Explanation:

Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:

Acceleration = (final velocity-initial velocity)/Time

Acceleration = (20-0)/10 = 2 m/s².

So this means the car is traveling with an acceleration of 2 m/s².

As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:

Time = Change in velocity/Acceleration

Time = (40-20)/2 = 20/2=10 s

So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.

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An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp
Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

k=\dfrac{1}{2}mv^2

k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2

k = 1150  J

In two significant figure, k=1.2\times 10^3\ J

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. P=\dfrac{W}{t}=\dfrac{\Delta K}{t}

P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}

P = 115 watts

In two significant figures, P=1.2\times 10^2\ Watts

Hence, this is the required solution.  

6 0
2 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi
AveGali [126]

Correct answer choice is:

C. Polarized in a vertical plane.

Explanation:

Polarized sunglasses give excellent glare shield, particularly on the water. Polarized lenses include a specific filter that prevents this type of strong reflected light, diminishing glare.

This is because when you angle one polarized lens to different perpendicularly, they prevent glare both horizontally and vertically. The polarized lenses are enduringly tinted sunglasses that exceedingly decrease glare.

7 0
3 years ago
Henry slides across an icy pond. The coefficient of kinetic friction betweenhis shoes and the ice is 0.09. If his mass is 115 kg
Kruka [31]

Answer:

101 N

Explanation:

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2 years ago
Most of the mass of the milky way exists in the form of.
Andrews [41]

Answer: Dark matter.

Explanation: Hope it helps :)

7 0
2 years ago
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