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Pepsi [2]
3 years ago
10

Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 6 seconds. What is the magnitude of the acceleration, includin

g units
Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The magnitude of the acceleration is 4.2 m/s²

Explanation:

Given;

initial velocity of the powerful sports car, u = 0

final velocity of the powerful sports car, v = 25 m/s

time of motion, t = 6 seconds

Acceleration = Δv / Δt

Δv is change in velocity

Δt is change in time

The magnitude of the acceleration of the powerful sports car during the motion is given by;

a = \frac{v-u}{t}\\\\a = \frac{25-0}{6}\\\\a = 4.2 \ m/s^2

Therefore, the magnitude of the acceleration is 4.2 m/s²

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Oduvanchick [21]

Answer:

15.0 m/s^2

Explanation:

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3 years ago
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Which of the following states of matter is always a good conductor of electricity? solid, liquid, gas, or plasma **:/ Thank you!
pogonyaev

Plasma...I believe is always a good conductor of electricity. I was tempted to say a solid, but not all solids are the same in composition and that goes for liquid and gas as well.

Hopefully this helped and good luck.

8 0
3 years ago
According to the periodic table, which of the following elements has five energy levels
Alex Ar [27]
<span>Antimony I am pretty sure is one. </span>
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3 years ago
Consider a spherical capacitor with radius of the inner conducting sphere a and the outer shell b. The outer shell is grounded (
AleksAgata [21]

Answer:

Explanation:

The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.

The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)

6 0
3 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

7 0
3 years ago
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