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Valentin [98]
3 years ago
8

A teacher pushed a 98 newton desk across a floor for a distance of 5 meter he exerted a horizontal force of 20 newton for four s

econds
Physics
1 answer:
Rom4ik [11]3 years ago
3 0
Missing question: "how much work is done by the teacher?" (found on internet)

Solution:
The work done to move the desk across the floor is equal to
W=Fd
where F is the horizontal force applied to move the desk, and d is the distance covered by the desk. If we use F=20 N and d=5 m, we find the work done:
W=Fd=(20 N)(5 m)=100 J
You might be interested in
A machine can never be 100% efficient because some work is always lost due to .
klasskru [66]
Friction
Hope it helped
5 0
3 years ago
Read 2 more answers
A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju
igor_vitrenko [27]

Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the  upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t      (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

8 0
3 years ago
A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-
wariber [46]

Answer:

Part a)

v = 12.45 m/s

Part B)

F_n = 0.05 N

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

mgH = \frac{1}{2}mv^2 + mg(2R)

so the speed at point A is given as

mg(H - 2R) = \frac{1}{2}mv^2

v = \sqrt{2g(H - 2R)}

v = \sqrt{2(9.81)(21.9 - 2\times 7)}

v = 12.45 m/s

Part B)

Now the force equation at point A is given as

F_n + mg = \frac{mv^2}{R}

F_n = \frac{mv^2}{R} - mg[/tex]

F_n = 0.004(\frac{12.45^2}{7} - 9.81)

F_n = 0.05 N

6 0
3 years ago
Why does the moon appear in the daytime
Brut [27]

Answer:

You can see the moon when it is in the perfect spot and it's reflecting the right amount of light. :)

Hope this helps! If you don't mind, please mark as brainliest! Thx!

4 0
2 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
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