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sveticcg [70]
3 years ago
5

What is the minimum force required to increase the energy of a car by 84 J over a distance of 38 m? Assume the force is constant

.
Group of answer choices

84 N

2.2 N

3200 N

120 N

0.45 N
Physics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

2.210N

Explanation:

Workdone = Force x distance

Distance = 38m , Workdone = 84J

Hence 84J = Force x 38m

Force = 84J / 38m

Force = 2.210N =2.2N

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1. A ball is thrown downward with an initial speed of 22 m/s on Earth. a. What is the acceleration of the ball? b. Calculate the
vitfil [10]

Answer:

Explanation:

The acceleration of the ball would be due to the downward force of gravity, 9.8m/s^2. In order to find the displacement given that interval of time, you have to use the corresponding kinematic formula:

d=v_it+at^2/2

The initial velocity was given, the time was given, and the acceleration was given. Therefore:

d=(22m/s)(4s)+(9.8m/s^2)(4s)^2/2

d=166.4m

To find the required time given a desired final velocity, we can use:

v_f=v_i+at

60m/s=22m/s+(9.8m/s^2)(t)

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2 years ago
When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted p
jarptica [38.1K]

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

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