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solniwko [45]
2 years ago
14

In which situation is the speed of the car constant while its velocity is changing?

Physics
1 answer:
Sati [7]2 years ago
7 0

Answer:

B, the car travels around a circular track at 30 m.

Explanation:

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The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
2 years ago
As part
umka21 [38]

Answer:

Part a)

a = 3.68 m/s^2

Part b)

a = 11.8 m/s^2

Explanation:

Part a)

For force conditions of two blocks we will have

m_1g - T = m_1 a

T - m_2g = m_2 a

now from above equations we have

(m_1 - m_2) g = (m_1 + m_2) a

a = \frac{m_1 - m_2}{m_1 + m_2} g

now we know that

m_1 = \frac{908}{9.8} = 92.65 kg

m_2 = \frac{412}{9.8} = 42 kg

now from above equation we have

a = \frac{92.65 - 42}{92.65 + 42}(9.8)

a = 3.68 m/s^2

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

F - mg = ma

908 - 412 = 42 a

a = 11.8 m/s^2

8 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
A 1500-kg car accelerates from 0 to 25 m/s in 7.0 s. What is the average power delivered by the engine (1 hp
LiRa [457]
Answer:

The average power delivered by the engine is 90 hp.
5 0
3 years ago
Suppose you tried to determine where we are in our galaxy by looking in different directions to see how many stars you could see
Olenka [21]

I don't know where you live, but chances are you won't see much at all. Light pollution levels have increased so much due to street lamps and other lights that it's practically impossible to see stars in the night time with the naked eye.

3 0
3 years ago
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