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gizmo_the_mogwai [7]
3 years ago
5

If an automobile engine delivers 42.0 hp of power, how much time will it take for the engine to do 6.20 â 105 j of work? (hint:

note that one horsepower, 1 hp, is equal to 746 watts.)
Physics
1 answer:
Elena L [17]3 years ago
7 0
To be able to answer this item, we are to calculate the power that the machine could deliver from hp to kW. 

      (45 hp)(746 W/1 hp) = 33570 W

Power is the amount of energy delivered at a certain period. 

             t = (6.20 x 10^2 J)/ (33570 kJ/s)

             t = 0.01845 s
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Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat
Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating  (assume it goes upto  h1 height )

using motion equations upwards

s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s  

2) motion under gravity (assume it goes upto  h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

                            = 15.625 + 7.8125

                            = 23. 4375 m

3 0
3 years ago
An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0
3 years ago
If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l
kakasveta [241]

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

5 0
3 years ago
Name and describe the two layers inside Earth that are undergoing convection and indicate one major consequence of each.
Lilit [14]

Answer:

Mantle and core

Explanation:

The Mantle and Core are the two components within Earth experiencing convection. In several ways the mantle is significant. The one outcome of convective current is the creation of the fresh oceanic lithosphere around OCEANIC RIDGES, formed by mantle upwelling. Core is indeed the planet's innermost layer.

5 0
3 years ago
Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer
Reptile [31]

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

I' = I_{0}Cos^{2}\theta     .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

I'' = I'Cos^{2}\left ( 90-\theta \right )

So,

I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )

I'' = I_{0}Cos^{2}\theta Sin^{2}\theta

I'' = \frac{I_{0}}{4}Sin^{2}2\theta

(b)

Now differentiate with respect to θ.

I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta

I'' = \frac{I_{0}}{2}\times Sin 4\theta

7 0
3 years ago
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