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Paul [167]
3 years ago
8

A car has a mass of 1,000 kilograms. The weight of the car is

Physics
1 answer:
ahrayia [7]3 years ago
6 0

Answer : The weight of the car is 9800 N

Explanation :

It is given that,

The mass of the car, m = 1000 kg

We have to find the weight of the car. The weight of an object is defined as the force acting on it under the action of gravity. The mass of an object is the amount of matter present in it.

Using second law of motion :

W = m g

Where

m is the mass of the car

g is the acceleration due to gravity g = 9.8 N/kg

W = 1000 kg × 9.8 N/kg

W = 9800 N

So, the weight of the car is 9800 N.

Hence, this is the required solution.

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Aneli [31]

Answer:

B.electromagnetic spectrum

Explanation:

Electromagnetic spectrum is used to  describe all of the wavelengths of light waves.Generally wavelength of the wave is denoted by λ.

Wavelength is the distance between two consecutive crest .Wavelength of the red color is minimum and wavelength of the violet color is maximum in the light  spectrum.

So the answer is B.

Electromagnetic spectrum

3 0
3 years ago
5 points
olga2289 [7]

Answer:

d. 5 ohms

Explanation:

For resistors in parallel, the equivalent resistance is found with:

1/Req = ∑(1/R)

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

R = 5

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3 years ago
What percentage of a lower trophic level's energy flows to the next higher trophic level? A. 1% b. 10% c. 50% d. 100% Please sel
ikadub [295]

Answer:

The answer is B; 10%

8 0
3 years ago
How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
densk [106]

Answer:

0.143 m

Explanation:

The relationship between force applied on a string and stretching of the spring is given by Hooke's law:

F=kx

where

F is the force exerted on the spring

k is the spring constant of the spring

x is the stretching of the spring from its equilibrium position

In this problem, we have:

F = 20 N is the force applied on the spring

k = 140 N/m is the spring constant

Solving for x, we find how far the spring will stretch:

x=\frac{F}{k}=\frac{20}{140}=0.143 m

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Mashutka [201]
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