Which is an action that can cause an object to move or change its state of motion

If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on

VMariaS [17]

The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>

8 0

2 years ago

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You travel 85 miles in 2 hours, you then travel 75 miles in 1.5 hours. After a 0.5 hour rest break you then travel 55 miles in 1

Lera25 [3.4K]

Total 215 miles in 5 hours .hope it helps

8 0

3 years ago

What crisis is occurring in California?

nlexa [21]

Hey there!

Option A

A climate crisis is occurring in California where sudden forest fires occur due to this .

3 0

2 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

2 years ago

The wavelength of visible light is about 5500 times longer than the wavelength of x-rays. If you represent the wavelenth of an x

Pie

Answer:

458.33 ft

Explanation:

We are given that

Wavelength of of an x- ray photon=1 in

Wavelength of of an x- ray photon== $\frac{1}{12}$ ft

1 in= $\frac{1}{12}$ feet

We have to find the length of line (in feet) drawn by you to represent the wavelength of visible light.

According to question

Wavelength of visible light= $5500\times \frac{1}{12}$ ft

Wavelength of visible light $=\frac{5500}{12}$ ft

Wavelength of visible light=458.33 ft

Hence, the 458.33 ft line must drawn by you to represent the wavelength of visible light.

6 0

3 years ago