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3 years ago

NEED HELP TODAY

Physics

2 answers:

larisa86 [58]3 years ago

7 0

Answer:

B- It is a constant when the light is traveling in a vacuum.

Explanation:

"Light in a vacuum always travels at the same speed." Those additional three words in a vacuum are very important. A vacuum is a region with no matter in it. So a vacuum would not contain any dust particles (unlike a vacuum cleaner, which is generally full of dust particles)

Free_Kalibri [48]3 years ago

6 0

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

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You are loading a refrigerator weighing 2267 N onto a truck, using a wheeled cart. The refrigerator is raised 1.09 m to the truc

meriva

Answer:

a).

Wmin= 2471.03 J

W=1603.01 J

Explanation:

Weight, w= 2267 N

$w= m*g\\W=m*g*h \\W=w*h$

Minimum work 'h' is the distance the refrigerator is raised h=1.09m

$W_{min}=2267 N* 1.09m\\W_{min}=2471.03 J$

The motion is no frictional force so, the magnitude of the force with a angle of 45.0° is find using:

$W=m*g*h'\\h'= sin(45)\\W=2267N*sin(45)\\W=1603.01 J$

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2 years ago

A skateboarder wants to cross a large playground and notices that there are large shapes painted on its asphalt surface. One sha

Elden [556K]

<h2>Diagonal of circle </h2>

Explanation:

As the skateboarder wants to cross the play ground . The surface is rough .

As we know , the force of friction is non-conservative force . Thus work is required against this force .

We have formula:

work done = Force x distance (in one direction )

Te force applied cannot be changed , so he is to decrease the distance .

In case of circle , diameter is the minimum distance . Thus he is supposed to move along it .

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3 years ago

a Ferrari with an initial velocity of 10 m/s, comes to a complete stop in 5 seconds, what will its acceleration be?

11111nata11111 [884]

10 m/s divided by 5 s

-2m/s/s

minus - deceleration

6 0

3 years ago

Read 2 more answers

What is the formula for wheel and axel

aleksley [76]

Answer: MA= Radius of wheel divided by Radius of Axle

Explanation:

3 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago