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3 years ago

State the average value of acceleration due to gravity

1 answer:

8 0

The acceleration due to gravity varies in different positions on Earth. However it is said to be negligible and 9.8 m/s is used. Sometimes people use 10 m/s or 9.81 m/s

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If you are standing on a weighing scale in an elevator what happens to your weight if the elevator accelerates up or accelerates

boyakko [2]

When moving upwards, the normal force from the weighing scale on the person increases so the weight appears to increase. The opposite holds true when the elevator goes downwards.

7 0

3 years ago

approximately what force,FM must the extensor muscle in the upper arm exert on the lower arm to hold a 7.6kg shot put? assume th

MAXImum [283]

Lets do the sum of the forces about the elbow joint.

Fm = Force of Muscle; Fe = Force Elbow; Fb = Force Ball

Sum Force about Joint = (-2.5)Fm + 12.5Fe + 30Fb = 0

(-2.5)Fm + 12.5(2.8) + 30(6.9) = 0

Fm = 96.8kg

Fm = 96.8 * 9.8 = 948.6N

Do you understand why the -2.5 is negative?
<span> Because I put the origin at the joint. So when you go left it is negative and when you go right it is positive. </span>

7 0

3 years ago

A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of

Burka [1]

Answer:

0.8m/s

Explanation:

Weight of mas,F=763 N

Mass of man= $\frac{F}{g}=\frac{763}{9.8}=77.86 kg$

By using $g=9.8m/s^2$

Weight of flatcar=F'=3513 N

Mass of flatcar= $\frac{3513}{9.8}=358.5 Kg$

Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg

Velocity of system=19.8m/s

Let v be the velocity of flatcar with respect to ground

Velocity of man relative to the flatcar= $-4.68m/s$

Final velocity of man with respect to ground=v-4.68

By using law of conservation of momentum

Initial momentum=Momentum of car+momentum of flatcar

$436.36(19.8)=77.86(v-4.68)+358.5v$

$8639.928=77.86v-364.3848+358.5v$

$8639.928+364.3848=436.36 v$

$9004.3128=436.36v$

$v=\frac{9004.3128}{436.36}$

$v=20.6 m/s$

Initial speed of flatcar=Speed of system

Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s

4 0

3 years ago

An air track car with a mass of 6 kg and velocity of 4 m/s to the right collides with a 3 kg car moving to the left with a veloc

sammy [17]

Remember that moment before collision is equal to the moment after collision.

$(m1 \times u1) + (m2 \times u2) = (m1 \times v1) + (m2 \times u2)$

Plugging in our values,

$(6 \times 4) - (3 \times 2) = (6 \times 1) + (3 \times v2) \\ 24 - 6 = 6 \times 3v2 \\ 18 = 18v2 \\ v2 = 1 {ms}^{ - 1}$

8 0

2 years ago

A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en

Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient)

W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)

This will then give us:

1/2Vi^2-ugd = 1/2V^2

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s

6 0

3 years ago