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3 years ago

7

A person gets a running start toward a wall from 5 meters away. The person pushes the wall with a force of 500 newtons, but the

wall does not move. How much work is done by the person?(1 point)
100 joules

0 joules

2,500 joules

0.01 joule

2 answers:

Ostrovityanka [42]3 years ago

6 0

100 joules is the answer.

yanalaym [24]3 years ago

4 0

Answer:

it's 0

Explanation: just took it

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A conductor bent into a semicircle of total length 3.88 m is placed in a 2.0 T magnetic field perpendicular to the plane of the

ivolga24 [154]

Answer:

Magnetic force, F = 38.95 N

Explanation:

Given that,

Length of the conductor, L = 3.88 m

Magnetic field, B = 2 T

Current flowing in the conductor, I = 5.02 A

The magnetic field perpendicular to the plane of the semicircle. The angle between the magnetic field and the plane is 90 degrees. The expression for the magnetic force is given by:

$F=ILB\ sin\theta$

$F=ILB$

$F=5.02\times 3.88\times 2$

F = 38.95 N

So, the magnitude of the total force acting on the conductor in the magnetic field is 38.95 N. Hence, this is the required solution.

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4 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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4 years ago

While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.

alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

Fg = m*g (always downward)
N (normal force, always perpendicular to the surface of the slide. going upward)
Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0

3 years ago

John sees Linda Running towards him at 11 m/s. while running, Linda throws a ball at 5m/s. what is the speed of the ball as obse

Ilia_Sergeevich [38]

11m/s Bc of the fact that he sees her running at 11m/s

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3 years ago

Crystals are made using supersaturated solutions of solids in water. These solutions hold more of the solute than they would at

Ilia_Sergeevich [38]

No 4 is the right answer.

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3 years ago