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DedPeter [7]
2 years ago
7

Question 2

Physics
1 answer:
DaniilM [7]2 years ago
4 0

P=m.v

v=P/m

v=500/20

v=25 m/s

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Why do you think we cannot use a common fuel like diesel or gasoline for space travel ?
Ede4ka [16]

Answer:

Explanation:

it take oxygen in the atmosphere to burn it... in space there isn't any air  :0

5 0
3 years ago
A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball
-Dominant- [34]

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 40 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

   4.9t²  + 2t - 40 = 0

   t = 2.66 s   or t = -3.06 s

So, time is 2.66 s.

Option C is the correct answer.

6 0
3 years ago
Two resistors, R1 and R2, are
dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)

1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)

<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0
3 years ago
: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature
Hunter-Best [27]

Answer:

The value is   \eta  =  0.2145  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  T_s =  100^o C =  100 + 273 = 373 \ K  

    The  second reservoir is at room temperature T_r  =  20^o C = 293 \ K

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     \eta  =  1- \frac{T_r}{T_s}

=>    1 - \frac{ 293}{373}

=>    \eta  =  0.2145

5 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
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