Question

Jupiter has radius pf 11 x the radius of the eart and a mass that is 320x the mass of the earth the gravitational field strength on the surface of jupiter is

GEarth =9.8ms^-2

A 3Nkg^-1
B 300 NG^-1
C 26 NG^-1
D 10 Nkg -1

An object of mass m at the end of a staring if length r moves in a vertical circle at a concentration angle speed w what is tension in the sting when the object is at the bottom of the circle

An object of mass m love horizontal circle of radio ur with constant speed what is the rate at which works is down by the centripetal force

Answer 1

C: 26 NG^-1

Part 2:

The rate at which work is done by the centripetal force is proportional to the cube of the velocity of the object.

Step-by-step explanation:

The gravitational field strength on the surface of Jupiter can be calculated using the formula:

gJupiter = G×MJupiter / rJupiter²

where G is the universal gravitational constant, MJupiter is the mass of Jupiter, and rJupiter is the radius of Jupiter. Using the given values, we get:

gJupiter = (6.67 × 10-11 N m2 kg-2) × (320 × MEarth) / (11 × REarth)2

gJupiter = 26.0 N kg-1

Therefore, the answer is option C.

For the second question, when the object is at the bottom of the circle, the tension in the string is equal to the weight of the object plus the centripetal force required to keep it moving in the circular path. The centripetal force is given by:

Fc = mv2 / r

where m is the mass of the object, v is the velocity of the object, and r is the radius of the circle.

At the bottom of the circle, the velocity of the object is maximum and equal to the square root of the product of the centripetal force and the radius divided by the mass of the object:

v = sqrt(Fc × r / m)

Substituting the value of Fc in terms of v and solving for tension T, we get:

T = mg + mv2 / r

T = m(g + v2/ r)

For the third question, the rate at which work is done by the centripetal force is given by:

P = Fc × v

where P is the power, Fc is the centripetal force, and v is the velocity of the object. Substituting the value of Fc in terms of v, we get:

P = mv3 / r

Therefore, the rate at which work is done by the centripetal force is proportional to the cube of the velocity of the object.

Answer 2

Step-by-step explanation:

Well this is quite tricky, as the gravitational field strength on the surface of Jupiter can be calculated using the formula:

g = G*M / r^2

Where G is the gravitational constant, M is the mass of Jupiter, and r is the radius of Jupiter.

Given that the radius of Jupiter is 11 times that of Earth (rJ = 11rE) and the mass of Jupiter is 320 times that of Earth (MJ = 320ME), we can substitute these values into the formula:

g = G x MJ / rJ^2

= G x (320ME) / (11rE)^2

= (G x 320 x ME) / (121 x rE^2)

Now, we know that G = 6.67 x 10^-11 N m^2 / kg^2 and gE = 9.8 m/s^2. So we can substitute these values and simplify:

g = (6.67 x 10^-11 N m^2 / kg^2 * 320 x ME) / (121 x rE^2)

= (2.14 x 10^16 N x ME) / rE^2

To get the gravitational field strength on the surface of Jupiter in terms of gE, we can divide g by gE:

g / gE = (2.14 x 10^16 N x ME) / (rE^2 x 9.8 m/s^2)

= (2.14 x 10^16 N x 5.97 x 10^24 kg) / ( (11 x 6.37 x 10^6 m)^2 x 9.8 m/s^2)

= 25.93

Therefore, the gravitational field strength on the surface of Jupiter is 25.93 times that of Earth.

Answer: C) 26 NG^-1

For an object of mass m at the end of a string of length r moving in a vertical circle at a constant angular speed w, the tension in the string at the bottom of the circle can be found using the formula:

T = mg + mv^2 / r

where g is the acceleration due to gravity, v is the velocity of the object at the bottom of the circle, and m is the mass of the object.

At the bottom of the circle, the object is moving horizontally, so the tension in the string is equal to the centripetal force required to keep it moving in a circle. The velocity of the object at the bottom of the circle can be found using the formula:

v = wr

where w is the angular speed of the object.

Substituting these values into the formula for tension, we get:

T = mg + m(wr)^2 / r

= mg + mw^2r

Therefore, the tension in the string at the bottom of the circle is T = mg + mw^2r.

Answer: T = mg + mw^2r

For an object of mass m moving in a horizontal circle of radius r with a constant speed v, the rate at which work is done by the centripetal force can be found using the formula:

W = Fc x v

where Fc is the centripetal force required to keep the object moving in a circle.

The centripetal force can be found using the formula:

Fc = mv^2 / r

Substituting this value into the formula for work, we get:

W = (mv^2 / r) x v

= mv^3 / r

Therefore, the rate at which work is done by the centripetal force is W = mv^3 / r.

Answer: W = mv^3