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2 years ago

What must be the reaction force when someone hits a tree with an axe?

Physics

1 answer:

Iteru [2.4K]2 years ago

7 0

Answer: The correct option is that the axe handle applying a force to the person's hand.

Explanation:

This reaction force is due to Newton's third law of motion. This law states that for every action there is equal and opposite reaction to it. This implies that when a force is being exerted on a surface, the surface must exert a force that is equal and opposite in direction to the exerting force. This law represents a certain symmetry in nature that forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

A typical example of Newton's third law includes:

--> When you hit a tree with an axe: with your hand you exert a force on the tree with the intention to cut it down(action force). The same force you excreted on the tree would be felt in your hands (reaction force).

--> when air rushes out of a balloon: when air escapes for a balloon, the opposite reaction is that the balloon flies up.

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2 things that can cause power output to decrease

Ad libitum [116K]

Answer:

idk and idk

Explanation:

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3 0

3 years ago

Read 2 more answers

A bottle rocket takes off with a = 34.5 m/s2. It

sergiy2304 [10]

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

Take them square root of both sides

t=√(11.59)

t=3.4 seconds

8 0

3 years ago

Medical x-rays, with a wavelength of about 10^(-10) m, can penetrate completely through your skin.

natulia [17]

Wavelength of X-rays = 10⁻¹⁰ m
Wavelength of UV = 1000 x 10⁻¹⁰
= 10⁻⁷ m

4 0

2 years ago

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 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$

$(6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)$

$v = (F_{xnet} - F_{x1}) / (F_{x2} )$

$=((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})$

$= (- 104.25) m/s$

$v_{y} = -104 m/s$

8 0

3 years ago