What happens when someone drops a watermelon from 25 storiea and a feather from 25 stories

Difference between weightlessness in space and weightlessness in earth

nadya68 [22]

In space we feel weightlessness because the earths gravity has less effect on us.The Earths gravitational attraction at those altitudes is only about 11% less than it is at the Earths surface. If you had a ladder that could reach as high as the the shuttles orbit, your weight would be 11% less at the top.

8 0

3 years ago

An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,

trasher [3.6K]

Answer:

The acceleration due to gravity is $\dfrac{1}{4}$ times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

$F=mg$

$mg=\dfrac{GmM}{R^2}$ ....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

$F'=mg'$

$mg'=\dfrac{GmM}{4R^2}$ ....(II)

Dividing equation (II) by equation (I)

$\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}$

$\dfrac{g'}{g}=\dfrac{1}{4}$

Hence, The acceleration due to gravity is $\dfrac{1}{4}$ times the value of g at the Earth’s surface.

4 0

4 years ago

Which of the following shows kinetic energy being converted into potential energy?

Komok [63]

B) a rock being tossed high into the air

3 0

3 years ago

In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

Fynjy0 [20]

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

8 0

3 years ago

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

3 years ago