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matrenka [14]
3 years ago
15

How long does it take a 100 kg rocket to increase speed from 10

Physics
1 answer:
oee [108]3 years ago
5 0

Answer:

76 seconds

Explanation:

Using the formula t=vf-vo/a, where t=time, vf= final velocity, v0= initial velocity, and a= acceleration, we can solve for your problem.

Plug in known variables...

t=200-10/a

we can solve for a by using the formula F=ma where f=force, m= mass, and a=acceleration and rearranging it to solve for a.

a=F/m

Plug in known variables...

a=250N/100kg

a=2.5m/s^2

Plug in our newly found acceleration of 2.5 m/s^2 into our first equation and solve....

t=190/2.5

t=76 seconds

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A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat
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Answer:

the  heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

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solving

Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J

the  heat absorbed by the block of copper is 74368.476J

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3 years ago
Explain whether a tennis ball dropped from a high distance experiences an elastic collision or inelastic collision
hodyreva [135]
I need help on that too poop head
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2 years ago
A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
BARSIC [14]

Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

E_i -E_f=W_f

\frac{1}{2}Kx^2=NU_kd

where K is the spring constant, x is the spring compression, N is the normal force of the block, U_k is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)

solving for U_k:

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8 0
3 years ago
The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se
Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

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Now, we can replace T1 from the first equation into the second one:

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Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

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X = 68s/0.094s/km = 723.4 km

6 0
3 years ago
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