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lorasvet [3.4K]
3 years ago
9

If you carry out an experiment measuring the weight and mass of objects in one particular location on the earth, what relation w

ill you find between weight and mass in your measurements?
A.They are equal.
B.Weight is inversely proportional to mass.
C.Weight is directly proportional to mass.
Physics
1 answer:
Elina [12.6K]3 years ago
8 0
The answer is letter C.

Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.

The force of attraction between any two masses is called Newton's Law of Universal Gravitation:

G \frac{m_1m_2}{d^2}

G is simply a given constant.

If we're at the surface of Eath, m_1 refers to the mass of the Earth, m_2 to the mass of whatever is on the surface of Earth, and d to the radius of Earth. 

Normally, we define a constant g to be equal to G \frac{M}{r^2}; in which M is the mass of Earth and r the radius of earth; g happens to be around 9.8.

By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.

W=mg

As you can see, weight is directly proportional to mass, more mass implies more weight.
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Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
mel-nik [20]

Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

m(2v)+m(-vsin\phi )=2m(ucos\theta )

2ucos\theta =v(2-sin\theta )------1

Conserving momentum in x direction

mvcos\phi =2musin\theta-----2

squaring and adding 1 &2

(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2

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4u^2=5v^2-4v^2sin\phi

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A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
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Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

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\theta=10.47^{\circ}

Hence, this is the required solution.

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