All categories

2 years ago

State a situation in which force is applied on a body, but no work is done

2 answers:

8 0

If you press on your arm force is applied work done is if it moves.
One answer could be if I was to press my hand on a table.
Have a great day!

Yanka [14]2 years ago

3 0

If a force is applied but the object doesn't move, no work is done; if a force is applied and the object moves a distance d in a direction other than the direction of the force, less work is done than if the object moves a distance d in the direction of the applied force.

You might be interested in

A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.

slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor .

R = Wman - T

3 0

2 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

2 years ago

What is the difference between the number of electrons in an atom of selenium, Se, and the number of electrons in an atom of alu

Oxana [17]

Well, electrons can be converted into a atomic number so if SE atomic number is 34 that means it has 34 electrons. AI has a atomic number of 13 meaning it has 13 electrons.

So the difference is that SE has more electrons then AI.

Hope this helped. :D

4 0

2 years ago

Read 2 more answers

What determines how much induced current will flow through a conductor?

Vitek1552 [10]

Answer:

Rate of change of magnetic flux

Explanation:

The induced current is equal to the ratio of induced emf to the resistance of the conductor.

According to the Faraday's law of electromagnetic induction, the induced emf is proportional to the rate of change of magnetic flux.

7 0

2 years ago

Please help me , thank you

mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

5 0

2 years ago

State a situation in which force is applied on a body, but no work is done​

State a situation in which force is applied on a body, but no work is done