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ELEN [110]
3 years ago
12

When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rat

e of 2.5 m>s 1150 m>min, or 500 ft>min2. What fraction of the engine power is being used to make the airplane climb
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

0.2289

Explanation:

Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

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Assume: Moving to the right is positive. A(n) 7.7 g object moving to the right at 22 cm/s makes an elastic head-on collision wit
krek1111 [17]

Answer:

Explanation:

We shall apply the formula for velocity in case of elastic collision which is given below

v₁ = (m₁ - m₂)u₁ /  (m₁ +  m₂)  + 2m₂u₂ / (m₁ +  m₂)

m₁ and u₁ is mass and velocity of first object , m₂ and  u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.

Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 )  + 2 x 18 x - 14 / ( 7.7 + 18 )

= - 8.817 - 19.6

= - 28.4 cm / s

5 0
3 years ago
If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0
3 years ago
Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
3 years ago
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis
tangare [24]

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules

Work done

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N

The magnitude of the force that is exerted on the handle is 56.86153 N

5 0
3 years ago
What is the difference between muscular endurance and muscular strength? A B or C
Galina-37 [17]

Answer:

Muscular endurance is how many times you can move a weight without getting tired.

Muscular strength is the amount force you can put out.

7 0
3 years ago
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