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finlep [7]
3 years ago
13

If the cart is moving with a velocity of magnitude 29.11m/s at the top of the loop-de-loop, what must be the magnitude of the no

rmal force from track on the cart at the top?
Physics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

The Normal Force is zero.

Explanation:

The reason for the normal for to be zero is because no matter if the is a form of contact at the top, the is no force applied by the cart to loop-de-loop, and so vice versa.

If the cart was on the bottom, then the normal force would be equal to the weight of the cart, which would be te force the cart would be applying to the loo-de-loop, and therefore by newton's third law, the loop-de-loop would apply an equal force to the cart, which we would say is the normal force.

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The star Betelgeuse is 6.1 x 10^18 m away from Earth. How old is the light we see from that star when it reaches us? There are 3
castortr0y [4]

Answer:

635 years old

Explanation:

The light reaching the earth from the sun will travel at a speed called the speed of light, and this has a universal value of 3 × 10⁸ m/s. Bearing this in mind, let us calculate the age of the light reaching the Earth from the sun:

Distance of star from Earth = 6.1 × 10⁸m

Speed of light = 3 × 10⁸ m/s

We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

Distance = speed × time

6.1 × 10⁸ = 3 × 10⁸ × time

time = \frac{6.1 \times 10^{18}}{3 \times 10^8}

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

∴ The number of years travelled by the light from the star:

3.2\ \times 10^7\ seconds = 1\ year\\1\ second =  \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}

please note that:

2.03 × 10¹⁰ = 20300000000

3.2 × 10⁷ = 32000000

\therefore \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}\\= \frac{20300000000}{32000000} \\\\= \frac{20300}{32} \\= 634.347\ years\\

The closest answer in the option is 635 years, and we are short of this by some points due probably to approximations in the calculation.

8 0
3 years ago
Plz help me it is improtant
marishachu [46]
I think it is b cause I don’t think you do that
7 0
3 years ago
Does a battery produce dc or ac? does the generator at a commercial power station produce dc or ac?
Art [367]
1) dc
2) ac

Because my dad said so
8 0
3 years ago
When you release the mass, what do you observe about the energy?
anygoal [31]

Explanation:

Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.

3 0
3 years ago
PLEASE HELP (100)
morpeh [17]

Answer:

The work done will be W=55.27\: J

Explanation:

The work equation is given by:

W=F\cdot x

Where:

F is the force due to gravity (weight = mg)

x is the length of the ramp (3 m)

Now, the force acting here is the component of weight in the ramp direction, so it will be:

F_{x-direction}=mgsin(28)

Therefore, the work done will be:

W=mgsin(28)*3

W=4*9.81*sin(28)*3

W=55.27\: J

I hope it helps you!

3 0
3 years ago
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