What do you know about potential energy? And how do we use it?

1 answer:

7 0

Energy is the ability to do work or cause change. There are basically two main types of energy, kinetic and potential. Potential energy is energy that is stored. There are various types of stored, or potential energy. Chemical energy from a battery is a potential form of energy, elastic energy in a stretched rubber band is a form of potential energy, but the most commonly referred to form of potential energy in physics is that of gravitational potential energy. This is energy that is stored due to an object's position. It is dependent on the mass of the object, the height of the object above the ground or Earth, and the acceleration due to gravity.

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A car takes 8s to increase its velocity from 10m/s to 30m/s.what is its average acceleration ?

DENIUS [597]

Acceleration is 2.5 m/s^2V1-V0/t= a so 30 m/s - 10 m/s / 8s = 2.5 m/s^2

7 0

4 years ago

Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal

notsponge [240]

Answer:

the angular acceleration is 9.7 rad/ $s^{2}$

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

τ = torque

I = inertia

σ = acceleration

moment inertia of this rod, I

I = $\frac{1}{3} mL^{2}$

τ = F d, d = $\frac{L}{2}$ cosθ

τ = m g $\frac{L}{2}$ cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

= (m g $\frac{L}{2}$ cosθ)/( $\frac{1}{3} mL^{2}$ )

= 3gcosθ/2L

= 3 (9.8)cos 38°/(2 x 1.2)

= 9.7 rad/ $s^{2}$

3 0

3 years ago

A stationary police car emits a sound of frequency 1200 hz that bounces off a car on the highway and returns with a frequency of

Whitepunk [10]

First, since the echo frequency is greater, the car must be roaming to the police car.
This is for the reason that the car is "running over" its own wave fronts, creating the wavelengths tinier, and henceforth increasing the frequency.

The trick here is to understand that the relative speed between the cars is half of the speed of the echo for the reason that the echo is returning to the police car at the similar speed as the sound wave trips to the moving car.

It would also appear that the speed of sound has been expected to be 337 m/s

So if v is the moving car's velocity, the correct formula to use is:

2v / 337 = (1250/1200) - 1
2v = 14.04 m/s
v = 7.02 m/s