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andreyandreev [35.5K]
3 years ago
8

Which is at rest relative to a seated passenger in a moving car?

Physics
2 answers:
AnnyKZ [126]3 years ago
5 0
THAT'S EASY BRO! the answer is the Seat!
Alisiya [41]3 years ago
5 0

Answer:

Explanation:

When a person is sitting in a moving car, he is in motion relative to the ground but he is at rest relative to the car.

Everything in the car is at rest relative to the person and the person is also at rest relative to other things in the car.

Thus, everything in the car is at rest relative to the person which is sitting in the car.  

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A. A cook leaves a plastic spoon and a wooden spoon in a pot of hot water for the same length of time. The spoons have the same
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a) The one with the higher capacity will have smaller change in temperature

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-C1M1(t-T1)=C2M2(t-T2)

t=\frac{C_1M_1T_1+C_2M_2T_2}{C_1M_1+C_2M_2}=26 C.

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Do you think most Americans are in good psychological health?
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Can someone help me a bit on this? Will mark brainliest. ( no physical science option soooo)
Advocard [28]

Answer:

When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.

The corresponding relation for diffraction is

d sin(\theta) \approx \lambda,

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7 0
3 years ago
Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7
Elden [556K]

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

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