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Serggg [28]
2 years ago
14

A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.

0 meters per second. What is the magnitude of the total momentum of the two fruit?
Physics
2 answers:
Murrr4er [49]2 years ago
8 0

Answer:

Hi how are you doing today Jasmine

Nat2105 [25]2 years ago
7 0

Answer:

16kg*m/s

Explanation:

a)p=m*v=2*4=8kg-m/s

b)p=m*v=1*8=8kg-m/s

add them

=16kg-m/s

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What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006
Ipatiy [6.2K]

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

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2 years ago
How does an object's position and velocity change as the object accelerates?”
taurus [48]

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

7 0
3 years ago
what is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes
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What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

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Therefore, the magnification of an astronomical telescope is -30.83.

5 0
3 years ago
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