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Mademuasel [1]
3 years ago
7

What is the similarities of a furnace and the sun?

Physics
1 answer:
Svetlanka [38]3 years ago
6 0

Answer:

They are both extremely hot, they both produce a form of light, they both have/use fire(typically)

Explanation:

You might be interested in
What is potential energy?
iVinArrow [24]
energy associate with position or shape
5 0
4 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
4 years ago
A force of 100 N acts upward. Resolve this force into 2 components; one that acts 30º north of west and one that acts 60º north
GrogVix [38]

To resolve these forces we have to make use of the sines and cosines.

To resolve this force in 30 degree north of west, the answer will be

100*sin(30)

The answer will be 50N

Now to resolve the force acting 60 degree north of east

100* cos(60)

The answer will be 50N.

This also adds to the total force acting that is 50+50=100N. This is the way forces are resolved according to their specified angles.

5 0
3 years ago
Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler
ehidna [41]

A) Acceleration in part A: 6.1 m/s^2

B) Acceleration in part B: 2.7 m/s^2

C) Acceleration in part C: 1.5 m/s^2

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

\Delta t = 0.9 - 0 = 0.9 s

So the acceleration is

a=\frac{5.5-0}{0.9}=6.1 m/s^2

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

\Delta t = 2.4 - 0.9 = 1.5 s is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

a=\frac{9.5-5.5}{1.5}=2.7 m/s^2

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

\Delta t = 3.4 - 2.4 = 1 s is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

a=\frac{11-9.5}{1}=1.5 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?
12345 [234]
This would be called the law of action-reaction. This states that every action will have an equal and opposite reaction. The action in the example is pulling down on the rope. The opposite and equal reaction is the climers body moving upward. The same law can be applied to a rocket. The action is the engines pushing down and the reaction is the rocket going up. :D
4 0
3 years ago
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