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lorasvet [3.4K]
4 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]4 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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Leokris [45]

Answer:

The solar cells transfer light energy to thermal energy.

When the battery is being charged up, chemical energy is transferred to electrical energy.

The motor is designed to transfer potential energy to kinetic energy.

5 0
4 years ago
a slender rod of mass m and length l is released from rest in a horizontal position. what is the rod's angular velocity when it
Makovka662 [10]

Answer:

M g H / 2 = M g L / 2      initial potential energy of rod

I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2   kinetic energy attained by rod

M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

8 0
2 years ago
An external force of 5N is applied to a go-kart that is opposite it’s direction of travel for 5 seconds. What is the resulting m
Greeley [361]

Answer:

25N.s=25kgm/s

Explanation:

The resulting momentum in this case is equal to the impulse created by the force 5N during 5 seconds

ΔP=Force.time=F.t

    = 5x5=25kgm/s

5 0
3 years ago
A hawk flies in a horizontal arc of radius 12.0 m at constant speed 4.00 m/s. (a) Find its centripetal acceleration. (b) It cont
olya-2409 [2.1K]

Answer:

a) a_c= 1.33 m/s^2

b) a= 1.79 m/s²

   θ = 41.98⁰

Explanation:

arc radius  = 12 m

constant speed = 4.00 m/s

(a) centripetal acceleration

     a_c=\frac{v^2}{R}

     a_c=\frac{4^2}{12}

                  = 1.33 m/s²

(b) now we have given

        a_t= \ 1.20 m/s^2

        now,

         a=\sqrt{a^2_c+ a^2_t}

         a=\sqrt{1.33^2+ 1.20^2}

            a= 1.79 m/s²

 direction

\theta = tan^{-1}(\frac{a_t}{a_r} )

\theta = tan^{-1}(\frac{1.2}{1.33} )

     θ = 41.98⁰

5 0
3 years ago
Please show steps as to how to solve this problem <br> Thank you!
liubo4ka [24]

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1  * X2 =

62.3 gm* 980 cm/sec^2  * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y   will be out of the page

r X F  for F1 will be into the page so the torque must be negative

6 0
3 years ago
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