3 years ago

KINEMATICS: MOTION ALONG STRAIGHT LINE

Physics

1 answer:

Vinil7 [7]3 years ago

6 0

Explanation:

hope this helps, cheers!

You might be interested in

Kinetic energy is the energy of . It increases with and .Potential energy is the energy of or . It increases with , and

lorasvet [3.4K]

Answer:

k is the energy of motion it increase with four of load you carry

8 0

3 years ago

A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

5 0

2 years ago

Read 2 more answers

What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?

Liula [17]

Answer:

a)1.37 s

b)∞ ( Infinite)

Explanation:

Given that

L= 47 cm ( 1 m =100 cm)

L= 0.47 m

On the earth :

Acceleration due to gravity = g

We know that time period of the simple pendulum given as

$T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}$

Here

$g_{eff}= g$

Now by putting the values

$T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}$

T=1.37 s

Free falling elevator :

When elevator is falling freely then

$g_{eff}= 0$ ( This is case of weightless motion)

Therefore

$T=2\pi\sqrt{ \dfrac{L}{0}$

T=∞ (Infinite)

6 0

3 years ago

Drag and drop the words into the correct locations to label the layers of the atmosphere, with the lowest layer at the bottom. (

professor190 [17]

Answer:

The top layer is exosphere. The fourth is the Thermosphere. The middle layer is the mesophere. The second layer is the Stratospere. the lowest layer is the troposphere

Explanation:

4 0

3 years ago

Read 2 more answers

A spring with spring constant of 33 N/m is stretched 0.15 m from its equilibrium position. How much work must be done to stretch

Hoochie [10]

Work done is 0.442J

<u>Explanation:</u>

Given:

Spring constant, k = 33 N/m

Distance, x₁ = 0.15m

Additional distance, x₂ = 0.072 m

Total distance = 0.15 + 0.072 m

= 0.222 m

Work done, W = ?

We can calculate work done by the formula

$W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2$

On substituting the value we get:

$W = \frac{1}{2}k [(x_2)^2 - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J$

Therefore, work done is 0.442J

7 0

3 years ago