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3 years ago

15

how far can your little brother get if he can travel at 2.5 m/s and in 5 seconds you will discover that his gin has run out of p

aint?

1 answer:

irga5000 [103]3 years ago

5 0

12.5 meters if he was going 2.5 m/s for 5 seconds.

No idea about the second part of the question

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gizmo_the_mogwai [7]

Answer:

a) P1=100kpa

V1=6m³

V2=?

P2=50kpa

rearranging mathematically the expression for Boyle's law

V2=(P1V1)/P2=(100×6)/50=12m³

b) same apartment as in (a) but only the value of P2 changes

=> V2=(100×6)/40=15m³

Explanation:

since temperature is not changing we use Boyle's law. mathematically expressed as P1V1=P2V2

4 0

3 years ago

A little girl is going on the merry-go-round for the first time, and wants her 50kg mother to stand near to her on the ride 2.1m

aliya0001 [1]

Angular momemtum : mass * tangential speed * distance to the center = 50*2.1*3.6=37800 J.s

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3 years ago

Yub87654d HELP there is this person trying to kid,.,nap,./, kids so if you see an account called Hernyana report there questions

Morgarella [4.7K]

Answer:

what???

Explanation:

7 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

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3 years ago

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

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