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2 years ago

What is the acceleration of an 24 kg object that applies a force of 130N?

Physics

1 answer:

WITCHER [35]2 years ago

6 0

Answer:

To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.

Explanation:

Given ,

F = 130N

M = 24kg

A = ?

F = m× a

then ,

130N = 24kg ×a

a = 130/24 = 5 m/s.

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3 years ago

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

$Q = ms \Delta T$

$Q = 1*s*(250 - 20)$

$Q = 1*s*230$

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

$Q = ms'\Delta T$

$1*s*230 = 1* s' * \Delta T$

now we have

$\Delta T = (\frac{s}{s'})*230$

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0

3 years ago

Read 2 more answers

Please help me with this question

valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

4 0

3 years ago

A car is initially travelling at 12.1m/s It accelerates at a rate of -3.9m/s^2 for 7 seconds. What is it’s final velocity?

rosijanka [135]

It will be traveling in the reverse direction it was originally going at 15.2 m/s

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3 years ago

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

zlopas [31]

Answer:

330.5 m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

$h=\frac{v^2_isin^2\alpha }{2g}$

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5 m

8 0

3 years ago