All categories

3 years ago

A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N

Physics

1 answer:

algol133 years ago

7 0

The weight of the person is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 40kg, g = 9.81m/s²

Plug in and solve for W:

W = 40(9.81)

W = 390N

You might be interested in

Physics students use a spring scale to measure the weight of a piece of lead. The experiment was performed two times: once in th

Lady_Fox [76]

Answer:

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

Explanation:

Knowing that the density of lead is $11,3 g/cm^{3}$ and the volume, we can calculate the true weight of the piece of lead:

$weight_{lead}=\rho _{lead}*V_{lead}=11,3 g/cm^{3} *50 cm^{3} = 565 g$

When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.

When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.

Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.

$Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2} -50 g*9.8 m/s^{2})}{9,8m/s^{2} } = 515 g$

Now that we have the two measurements, we can calculate the difference:

$Difference= |Weight _{in water}- Weight _{in air}|=|515 g-565 g|=50 g$

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

4 0

3 years ago

The experimenter fom the video rotates on his stool, this time holding his empty hands in his lap. You stand on a desk above him

gtnhenbr [62]

Answer:

Explanation:

The experimenter is rotating on his stool with angular velocity ω ( suppose )

His moment of inertia is I say

We are applying no torque from outside . therefore , the angular momentum will remain the same

Thus angular momentum L = I ω = constant

Thus we can say I₁ ω₁ = I₂ω₂ = constant

here I₁ is the initial moment of inertia and ω₁ is the initial angular velocity

Similarly I₂ is the final moment of inertia and ω₂ is the final angular velocity

When a been bag is dropped on his lap , his moment of inertia increases due to increase in mass

In the above equation, when moment of inertia increases , the angular velocity decreases . So its motion of rotation will decrease .

7 0

3 years ago

How do land and sea breeze work?

Studentka2010 [4]

by the wind and air flow in the wind

7 0

4 years ago

At a local swimming pool, the diving board is elevated h = 9.5 m above the pool's surface and overhangs the pool edge by L = 2 m

eimsori [14]

Answer:

1) The time it takes the diver to move off the end of the diving board to the pool surface, $t_w$ , is approximately 1.392 seconds

2) The horizontal distance from the edge of the pool to where the diver enters the water, $d_w$ , is approximately 5.76 meters

Explanation:

1) The given parameters are;

The height of the diving board above the pool's surface, h = 9.5 m

The length by which the diving board over hangs the pool L = 2 m

The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s

Taking $t_w$ = The time it takes the diver to move off the end of the diving board to the pool surface

Therefore, we have from the equation of free fall;

h = 1/2 × g × $t_w$ ²

Where;

g = The acceleration due to gravity = 9.81 m/s²

Substituting the values, gives;

9.5 = 1/2 × 9.81 × $t_w$ ²

$t_w$ = √(9.5/(1/2 × 9.81)) ≈ 1.392 s

The time it takes the diver to move off the end of the diving board to the pool surface = $t_w$ ≈ 1.392 s

2) The horizontal distance, $d_w$ , in meters from the edge of the pool to where the diver enters the water is given as follows;

$d_w$ = L + v₀ × $t_w$ = 2 + 2.7× 1.392 ≈ 5.76 m

∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.

7 0

3 years ago

Is a puddle of water an object in motion?

Brilliant_brown [7]

Yes, a puddle of water is an object in motion.

4 0

3 years ago