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poizon [28]
3 years ago
11

What is the fundamental source of all energy in the universe

Physics
1 answer:
seropon [69]3 years ago
5 0
The Sun is the world's main source of energy. Without it we would not be able to live.
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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
4 years ago
How does the balloon react to the cloth item?
almond37 [142]

Answer: When rubbing a balloon with a wool cloth, it puts negative charges on the balloon. Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon.

8 0
3 years ago
When a certain amount of current flows through a resistor, it uses 3.00 W of power. If the current doubles, how much power will
asambeis [7]

Answer: 12

Explanation: Acellus

7 0
3 years ago
Okay i'm totally stuck and nobody I know really gets it either, so i've turned to Yahoo for help :)
OlgaM077 [116]

Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:

                     (weight) x (distance from the pivot) <u>on one side</u>
is equal to
                     (weight) x (distance from the pivot) <u>on the other side</u>.

That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.

       (Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).


So now we come to the strange beings on the alien planet.
There are three choices right away that both work:

<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.

       (400) x (1)  =    (200) x (2)

<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.

       (200) x (1)  =    (100) x (2)

<u>#3).</u>

On one side:  (300 N) in the end-seat       (300) x (2) = <u>600</u>

On the other side:
                      (400 N) in the middle-seat  (400) x (1) = 400
           and     (100 N) in the end-seat      (100) x (2) = 200
                                                    Total . . . . . . . . . . . . <u>600</u> 


These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.


5 0
3 years ago
Read 2 more answers
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
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