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3 years ago

A 39.7 n object is in free fall. what is the magnitude of the net force which acts on the object? answer in units of n.

1 answer:

6 0

If there is no air resistance, the value would still be the same. So therefore, the answer is 39.7 N.
Supposing that there is an air resistance of 19.8 N.With a 19.8 N air resistance: ∑F = 39.7 N - 19.8 N = 19.9 N would be the net force. The important point here is that the force of gravity equivalent the weight of the object.

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Compared to its weight on Earth, a 5kg object on the moon will weigh

shutvik [7]

Answer:

8.1 N/49 N=0.1653 which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 $\frac{m}{s^2}$ , the weight of the 5 kg object will be: $weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N$

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

$weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N$

Therefore the object's weight on the Moon compared to that on Earth will be:

$8.1N/49N=0.1653$

That is, 16.53% of the weight the object has on Earth.

5 0

3 years ago

A girl is whirling a ball on a string around her head in

Thepotemich [5.8K]

Answer:

Answered

Explanation:

The girl whirling the ball should let go off ball when the ball is at a position such that tangent to the circle is in the direction of the target.

the tangent at any point in a circular path indicates the direction of velocity at that point. And the moment when the centripetal force is removed the ball will follow the tangential path at that moment.

8 0

3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

Frictional Force

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

7 0

3 years ago

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$

here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

$F_{net} = 4501.9 N$

so it will have net force 4501.9 N which will be reported by sensor

4 0

3 years ago

Read 2 more answers

In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v

dangina [55]

Volume of gold in the phone = 10 cm^3
= 0.00001 m^3 
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0.00001 m^3 * (19300 kg/m^3)
= 0.193 kg
So
0.193 kg = 0.193 * 2.2 pounds
= 0.43 pounds
I think there is something wrong with the options given in the question.

7 0

3 years ago