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Crank
3 years ago
6

If you could throw an object high enough to reach space how hard would you have to throw it

Physics
1 answer:
natulia [17]3 years ago
3 0

To throw an object into the space you need to throw it with the escape velocity of the earth which is 11.2 km/sec. therefore if you can throw an object with the speed of 11.2 km/sec it will leave the earth atmosphere and will reach to the space.

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A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif
forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

5 0
3 years ago
A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its ra
mr_godi [17]
<h2>Answer:</h2>

The rate of deceleration is -0.14m/s^{2}

<h2>Explanation:</h2>

Using one of the equations of motion;

v = u + at

where;

v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)

u = initial velocity of the boat = 25m/s

a = acceleration of the boat

t = time taken for the boat to accelerate/decelerate from u to v =   3 minutes

<em>Convert the time t = 3 minutes to seconds;</em>

=> 3 minutes = 3 x 60 seconds = 180seconds.

<em>Substitute the values of v, u, t into the equation above. We have;</em>

v =  u + at

=> 0 = 25 + a(180)

=> 0 = 25 + 180a

<em>Make a the subject of the formula;</em>

=> 180a = 0 - 25

=> 180a = -25

=> a = -25/180

=> a = -0.14m/s^{2}

The negative value of a shows that the boat is decelerating.

Therefore, the rate of deceleration of the speed boat is 0.14m/s^{2}

5 0
3 years ago
Question 4 - If Angelica starts out at 30m/s, and in 16 s speeds up to 84 m/s, what is her acceleration?
Triss [41]

Answer:

C. 3.375 m/s^2

Explanation:

The acceleration of an object can be found using the equation:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

a=\frac{84-30}{16}=3.375 m/s^2

4 0
3 years ago
What is the mass of an atom with six protons, seven neutrons, and eight electrons?
qaws [65]

Answer:

13

Explanation:

The mass of an atom= number of protons + number of neutrons

7 0
3 years ago
A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
3 years ago
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