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antiseptic1488 [7]
3 years ago
8

Give an example where instantaneous speed is different from average speed

Physics
2 answers:
Ulleksa [173]3 years ago
6 0
The speedometer of a car reveals information about the instantaneous speed of your car.
It shows your speed at a particular instance in time...
Yet, on average, you were moving with a speed of 25 miles per hour.

Hope you liked the example, and glad I could help!
Brut [27]3 years ago
6 0

Answer:

Yes average speed may be different from instantaneous speed

Explanation:

When an object is moving with non uniform speed on a straight road or on a curved path then its instantaneous speed at different points of its path is different.

But if we will find the average speed of the path then it will be different from its actual average speed

For example we can say that let a Car is moving for speed 40 km/h for first hour of its journey and then for next hour of its journey its speed is 60 km/h

So here instantaneous speed of the car is either 40 km/h or 60 km/h

While to find the average speed we will say

v_{avg} = \frac{v_1 + v_2}{2}

v_{avg} = \frac{40 + 60}{2} = 50 km/h

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A moving car has momentum. if it moves twice as fast its momentum is ____________ as much
Nikolay [14]
It's momentum is twice as much.
3 0
3 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

4 0
3 years ago
If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will
densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

P=\frac{V^2}{R}

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P

so, the power dissipated will quadruple.

6 0
3 years ago
Which is not an element? Water , Arsenic, sodium, Aluminum
MrRa [10]

Answer:

water

Explanation:

water is not an element, it is a molecule

3 0
3 years ago
Read 2 more answers
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
3 years ago
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