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Mnenie [13.5K]
3 years ago
12

3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?

Physics
1 answer:
makkiz [27]3 years ago
7 0
I think it is B. 100 W
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Un comerciante compra 300 litros de aceite a 3 L y los vendió 5 L qué beneficio obtuvo​
andre [41]
El beneficio que obtuvo es de 2 L
5 0
3 years ago
Planet RMM-1 has a mass of 28,500 kg and the star it revolves around has
SOVA2 [1]

Gravitational force = G · (mass₁) · (mass₂) / (distance)

(distance²) = G · (mass₁) · (mass₂) / (Gravitational force)

G = 6.67 x 10⁻¹¹ n-m² / kg²  (the "gravitational constant")

Distance²  = (6.67 x 10⁻¹¹ n-m² / kg²) (28,500 kg) (2.2 x 10⁸ kg) / (39 N)

Distance² = (6.67 · 28,500 · 2.2 x 10⁻³ N-m²) / (39N)

Distance²  =  (418.209 N-m²) / (39N)

Distance²  =  10.72 m²

<em>Distance = 3.275 meters</em>

An absurd scenario, but that's by golly what the math says with the numbers provided.  I guess it's a teeny tiny planet orbiting 3.275 meters outside a teeny tiny black hole.

3 0
3 years ago
A pilot took 1/2 hour to complete 1/3 of his trip. How much longer will it take him to complete his trip?
katen-ka-za [31]

Answer: The answer is 1.5 hours.


Step-by-step explanation: Given that the pilot took 1/2 hour to complete 1/3rd of his trip. We are to calculate the time that he will take to complete his trip.

Time taken by the pilot to complete 1/3 rd of his trip = 1/2

Therefore, time taken by the pilot to complete his trip is given by

t=\dfrac{1}{2}\times 3=1.5.

Thus, the pilot will take 1.5 hours to complete his trip.

7 0
3 years ago
Is the force of gravity stronger on a piece of paper when it is crumpled? is the force of gravity stronger on a piece of iron th
Kipish [7]

No, gravity pulls every object with the same force irrespective of its shape, size place etc.

Therefore the force of gravity on the iron and wooden piece of same mass is same. Also the force of gravity is same on the crumpled piece of paper as that of plain paper.

Force of gravity is always same.

5 0
3 years ago
How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr
Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

Q =mc\Delta T

where,

m = mass

c = specific heat

\Delta T = Change in temperature

Therefore the total heat exchange is given as

\Delta Q = Q_w+Q_v

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

Our values are given as,

Total mass is M_T = 200lb ,however the mass of solid vegetable and water is given as,

m_v= 0.4*200lb = 80lb

m_w=0.6*200lb=120lb

T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF

Replacing at our equation we have,

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)

\Delta Q = 22411.2Btu

Therefore the heat removed is 22411.2 Btu

6 0
3 years ago
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