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3 years ago

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Physics

1 answer:

Korolek [52]3 years ago

6 0

Answer:

7.5 m/s2

Explanation:

F = ma

300 = 40 kg (a)

divide both sides by 40 to single out the variable a (acceleration)

300/40 = 7.5 m/s2

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This equation goes with which law?

Alchen [17]

Answer:

I think that's Newton's second law of motion

Explanation:

f = m(v-u)

________

since a = (v-u)t

f = ma

3 0

3 years ago

Will give correct answer brainliest

sineoko [7]

$200 \times 80\%$

<h2>calculate</h2>

$200 \times \frac{80}{100}$

<h2>reduce the numbers</h2>

$2 \times 80$

<h2>multiply</h2>

$= 160$

<h2>there for we have a solution to the equation</h2>

hope it helps

#carry on learning

mark me as brainlist plss

6 0

3 years ago

Read 2 more answers

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ikadub [295]

In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

3 years ago

Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

shepuryov [24]

Answer:

b) TA = TB = TC

Explanation:

When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
The equilibrium condition will be reached when the following equation be met:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
So, option b) is the right one.

8 0

3 years ago

How much energy is required to raise the temperature of 5g of air by 10°C?

Alex777 [14]

You need to know the specific heat capacity of air.
Then energy needed = 0.005 x sp.heat.cap x 10

4 0

3 years ago