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12345 [234]
3 years ago
7

A mass of 7.5kg has a weight of 30n on a certain planet calculate the acceleration due to gravitt on this planet

Physics
1 answer:
Over [174]3 years ago
6 0

Answer:

Acceleration due to gravity on a certain planet = 4 m/s²

Explanation:

According to the question,

Weight = 30 N

Mass = 7.5 kg

Let acceleration due to gravity be 'a'

<u>Formula</u>:

Weight = Mass × Acceleration due to gravity

30 = 7.5 × a

a = 30/7.5

a = 4 m/s²

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A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of
Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

8 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

\displaystyle A=\frac{3}{2}B^mC^n

and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

m+n=2

-m+2n=2

Adding both equations:

3n=4

Solving:

n=4/3

m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
3 years ago
What is the relationship between the masses of the objects and the gravitational force between them
irga5000 [103]

The relationship between the masses of the object and the gravitational force between them is a direct relationship

Explanation:

The gravitational force between two objects is given by  the equation:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

We observe that:

- The gravitational force is proportional to the masses of the two objects, m1 and m2, so if the masses increase, the force will increase as well (so, this is a direct relationship)

- The gravitational force is inversely proportional to the square of the separation between the objects, so if the distance is increased, the force will decrease (so, this is an inverse relationship)

Learn more about gravitational force here:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
3 years ago
An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir
chubhunter [2.5K]
The photocell<span>-- The click rate depends upon the filter selected.</span>
8 0
3 years ago
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What is a break in rock? Science
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Its physical weathering and physical weathering can be sometimes called mechanical weathering it includes the processes which break rocks apart changing their chemical composition.
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