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Morgarella [4.7K]
3 years ago
7

If we took our laboratory spring to the moon, how accurately would it measure the mass? Why?

Physics
1 answer:
a_sh-v [17]3 years ago
6 0

Accurately

Explanation:

If we took our laboratory spring to the moon, we can accurately measure the mass because mass of any substance is the same every where in the universe.

Mass is the amount of substance in a matter. Since the same substance on earth is made up of same matter on moon, the mass will remain the same.

  • The only difference between the measurement would be the weight.
  • Weight is the vertical force on an object.
  • It is function of mass and acceleration due to gravity.

Weight  = mass x acceleration due to gravity.

  • Acceleration due to gravity on earth and moon are different.
  • The gravity on moon is about one-sixth that on earth.
  • The weight using the same mass will be one-sixth
  • But the mass will be the same.

learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

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One earth day equals 24 hours.
Artyom0805 [142]

Answer:

Correct, is there another part to the question?

7 0
3 years ago
Read 2 more answers
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
Name the net force between objects that allows a car to travel in a circle
iren [92.7K]
I think centripetal force ☺
6 0
3 years ago
In a lab, four balls have the same velocities but different masses.
olya-2409 [2.1K]

Answer:

New Momentum of Ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2\frac{m}{s}

Momentum of Ball A=2.2\frac{k g m}{s}

Momentum of Ball B=4.4 \frac{k g m}{s}

Momentum of Ball C=11\frac{k g m}{s}

Momentum of Ball D=15\frac{k g m}{s}

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2\frac{m}{s}

Initial Momentum of Ball B=4.4\frac{k g m}{s}

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

                                            =3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,  

Change in momentum of Ball B \Delta p=m \times \Delta v

Where \Delta p=change in momentum

          m=mass of the ball B

         \Delta v=change in velocity ball B

And \Delta v=v, since all balls, have same velocity

Thus the above equation, changes to

         \Delta p=m \times v

Substitute all the values in the above equation we get

         \Delta p=6 \times 2.2

                     =13.2 \frac{\mathrm{kgm}}{\mathrm{s}}  

Result:

 Thus the New Momentum of ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

3 0
3 years ago
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            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
7 0
3 years ago
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