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san4es73 [151]
1 year ago
13

According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine

2. air resistance

3. the friction between the road and the tires

4. gravity
Physics
1 answer:
Leni [432]1 year ago
6 0

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

W=\dfrac{1}{2}m(v^2-u^2)

Also, W = F.d

Fd=\dfrac{1}{2}m(v^2-u^2)

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

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Svetradugi [14.3K]

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

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2 years ago
A spaceship which is 50,000 kilometers from the center of Earth has a mass of 3,000 kilograms. What is the magnitude of the forc
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Fg=Gx(M1M2/r^2)
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6 0
1 year ago
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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

i_1+i_3=i_2                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

i_3=i_2-i_1=0.75A-0.40A=0.35A

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0
2 years ago
Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted
34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0
2 years ago
Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force
jarptica [38.1K]

Hi!


Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.


Hope this helps!


3 0
2 years ago
Read 2 more answers
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