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1 year ago

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When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

otential energy? It decreased by a factor of 3. It decreased by a factor of 2. It increased by a factor of 2. It increased by a factor of 3.

1 answer:

Gnesinka [82]1 year ago

6 0

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

You might be interested in

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

2 years ago

In order to get a tree stump out of the ground, chains are connected to two trucks. One truck pulls with a force of 600 N to the

Black_prince [1.1K]

Answer:

The net force on the stump is 1000 N.

Explanation:

Given that,

Force 1 acting on the truck, $F_1=600\ N$ (due north)

Force 2 acting on the truck, $F_2=800\ N$ (due west)

We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :

$F=\sqrt{F_1^2+F_2^2}$

$F=\sqrt{600^2+800^2}$

F = 1000 N

So, the net force on the stump is 1000 N. Hence, this is the required solution.

3 0

2 years ago

An uncharged spherical conducting shell surrounds a charge –q at the center of the shell. Then charge +3q is placed on the outsi

Veronika [31]

Answer:

a) The the charges on the inner and outer surfaces of the shell are respectively +q, -q

Explanation:

Under static equilibrium inside a conductor, the total electric field, E = 0

This must be zero so that no charge will be moving since the conductor is in static equilibrium.

Also, since Electric field, E is zero, then flux through the surface will zero.

From Gauss' law, the total charge enclosed is zero.

Given –q as the charge at the center of the shell, then the opposite charge on inner surfaces will be +q, so that the total charge enclosed will be zero.

Since the charge is in static equilibrium, then opposite charge will be on the surface, that is –q.

Therefore, the the charges on the inner and outer surfaces of the shell are respectively +q, -q

8 0

1 year ago

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

a)

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

b)

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

c)

The work is equal to

W = 5886 [J]

d)

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

1 year ago

An ideal gas has a density of 1.75 kg/m3 at a gauge pressure of 160 kPa. What must be the gauge pressure if a density of 1.0 kg/

Mashutka [201]

Have you tried search this up?

5 0

1 year ago

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