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Wittaler [7]
3 years ago
5

What is newton's second law of motion?

Physics
1 answer:
lord [1]3 years ago
3 0

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

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A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which of the following s
Ganezh [65]

Answer:

d. It is equal to the component of the gravitational force acting down the ramp.

Explanation:

The stationary crate is inclined at an angle with the horizontal. The Recall, Frictional Force is any Force that opposes motion.

Because the Force of Friction that is opposing the motion of the crate along the inclination side.

Therefore this Frictional force is balanced or equal to the force that is driving the inclined force.

Hence Frictional Force is equal to the Gravitational Force that is acting in the ramp, that is why the crate is stationery.

8 0
3 years ago
Thing 1 and Thing 2 push with a force of 2000 N to move a rather small 500 kg elephant 30 m across the kitchen floor. How much w
Amanda [17]

Answer:

60000 J

Explanation:

Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

W=Fd

where

F = 2000 N is the force applied

d = 30 m is the displacement of the elephant

Substituting the numbers into the formula, we find

W=(2000 N)(30 m)=60,000 J

3 0
3 years ago
a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0
3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
3 years ago
Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.
GalinKa [24]
Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.
7 0
3 years ago
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