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rusak2 [61]
2 years ago
15

Which of the following BEST describes what a thermometer measures

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

A theromometer is the increase or decrease of earths atmospheric temperture, thats how you would measure the temperture of the air around you.

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Hello, PLEASE ANSWER ASAP!!! Don't report my question or give me false answers but I'm freaking out. I was biking home from scho
kodGreya [7K]

Answer:

just calmly talk and get money to pay them the bike and explain it to them

Explanation:

4 0
3 years ago
A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a
Rama09 [41]

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

3 0
3 years ago
A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th
BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

s = ut + \frac{1}{2}at^2

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

156.3 = 31\times t + 0

t = \frac{156.3}{31 }

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 m/s^2

Substituting in h = ut + \frac{1}{2}gt^2

h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2

h = 124.694 m

So height of ramp = 124.694 m

3 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
2 years ago
A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the
Nat2105 [25]

Answer:

Explanation:

mdeemmkdkwdmwdmw

4 0
3 years ago
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