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3 years ago

What is the work done if you push a tree with 50 N of force but the tree does not move?

Physics

1 answer:

Soloha48 [4]3 years ago

4 0

Answer:

The work done is 0.

Explanation:

The reason no work is done is because the equation W = Fs.

W = work

F= force

s= displacement

In this scenario F = 50 and s= 0

Therefore.

W = 50(0)

W = 0

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A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh

otez555 [7]

Answer:

The percentage of the weight supported by the front wheel is A= 19.82 %

Explanation:

From the question we are told that

The center of gravity of the plane to its nose is $z = 2.58 m$

The distance of the front wheel of the plane to its nose is $l = 0.800\ m$

The distance of the main wheel of the plane to its nose is $e = 3.02 \ m$

At equilibrium the Torque about the nose of the airplane is mathematically represented as

$mg (z- l) - G_B *(e - l) = 0$

Where m is the mass of the airplane

$G_B$ is the weight of the airplane supported by the main wheel

$G_B =\frac{mg (z-l)}{(e - l)}$

Substituting values

$G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}$

$G_B = 0.8018 mg$

Now the weight supported at the frontal wheel is mathematically evaluated as

$G_F = mg - G_B$

Substituting values

$G_F = mg - 0.8018mg$

$G_F = (1 - 0.8018) mg$

$G_F = 0.1982 mg$

Now the weight of the airplane is = mg

Thus percentage of this weight supported by the front wheel is $A = 0. 1982 *100 =$ 19.82 %

7 0

3 years ago

A student does 25 J of work on the handle of a pencil sharpener. If the pencil sharpener does 20 J of work on the pencil, what i

allsm [11]

Answer:

80%

Explanation:

Efficiency of machine = work output/work input ×100 %

From question, work output = 20J

Work input = 25J.

Therefore efficiency = 20/25 × 100 %

Efficiency = 20×4 %

Efficiency = 80%

I hope this was helpful, please mark as brainliest

3 0

3 years ago

If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the

hoa [83]

Answer:

5.5 km

Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

= 252.78 m/s

Now, we use the formula v²/r = gtanθ to get our needed radius

making r the subject of the formula, we have

r = v²/gtanθ, where

r = radius of curvature needed

g = acceleration due to gravity

θ = angle of banking

r = 252.78² / (9.8 * tan 50)

r = 63897.73 / (9.8 * 1.19)

r = 63897.73 / 11.662

r = 5479 m or 5.5 km

Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km

4 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.

astra-53 [7]

Answer:

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0

3 years ago

In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-

pogonyaev

Answer:

a) truc is C, b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

v = d / t

v = 3 109 m / 20

v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time t = to γ

For distance L = Lo / γ

γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time t = 20 1.15 = 23 s

The distance L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0

3 years ago