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Rama09 [41]
1 year ago
14

Jordan wants to know the difference between using a 60-W and 100-W lightbulb in her lamp. She calculates the energy it would tak

e to use each bulb for 30 s by using this equation: Power = Energy transferred (J)/time (s) What is the difference in the amount of energy transferred by the two bulbs during this time?
Physics
1 answer:
Andreas93 [3]1 year ago
4 0

Answer:

The difference in the amount of energy transferred by the two bulbs is 1200 J.

Explanation:

The energy transferred by the two lightbulbs can be calculated with the given equation:

E = P*t

Where P is the power and t is the time

For the 60 W lightbulb:

E_{60} = P*t = 60 W*30 s = 1800 J

For the 100 W lightbulb:

E_{100} = P*t = 100 W*30 s = 3000 J

Hence, the difference in the amount of energy transferred is:

E_{t} = E_{100} - E_{60} = 3000 J - 1800 J = 1200 J

Therefore, the difference in the amount of energy transferred by the two bulbs is 1200 J.

I hope it helps you!

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A truck is using a hook to tow a car whose mass is one quarter that of the truck. If the force exerted by the truck on the car i
LenKa [72]

Answer:

Force exerted by car on truck will be 6000 N in opposite direction

Explanation:

It is given that mass of the car is one quarter of the mass of the truck

Force exerted by the truck on the car is 6000 N

We have to find the force exerted by car on the truck

According to newtons third law for any action there is equal and opposite reaction

So force exerted by car on the truck will be equal to 6000 N in opposite direction

7 0
1 year ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.
Dimas [21]

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

5 0
2 years ago
Can a small child play with fat child on the seesaw?Explain how?
RoseWind [281]
Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
8 0
1 year ago
Read 2 more answers
In a distant solar system, a giant planet has
sergeinik [125]

Answer:

mass of the planet: 5.9\,10^{26}\,kg

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

m\frac{v^2}{R}=G\frac{M\,m}{R^2}

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg

8 0
2 years ago
What are close-toed shoes least likely to provide protection against?
Sphinxa [80]

Hmm.. Well i would say A or C because water and acid would just soak threw it

7 0
1 year ago
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