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Effectus [21]
3 years ago
13

A car accelerates from rest at a constant rate 5m/s^2. Which of the following statements is true

Physics
1 answer:
dusya [7]3 years ago
8 0

Answer:

jjhbkvjuhigjgyihbgtimnvyuoibc

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a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0
3 years ago
In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.
Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

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

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

g=9.8 m/s^2 (acceleration of gravity)

Therefore the period of this pendulum is

T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s

#LearnWithBrainly

3 0
3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
Help please !!<br>I need help with this!
belka [17]
Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m
6 0
2 years ago
A slide with an image 4cm×2cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen
pav-90 [236]

Answer:

44cm x 22cm

Explanation:

u= 10 cm

v= 1.1 cm

m=v/u= 1.1/10

m=11

hence the size of the image.

7 0
2 years ago
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