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Oksanka [162]
3 years ago
11

What is the G.P.E of 675 newtons on 26000 meter mountain

Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0
GPE=mgh, where m=mass, g=acceleration due to gravity, h=height
We have a force of 675N already, so mg=675N

<span>GPE=mgh=675(26000)=17.55x10^6J</span>
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What can you predict about an element from its position in the periodic table
bazaltina [42]
Based on its position in the periodic table, you can predict how many electrons it has, how many valence electrons, how many levels of electrons, and its atomic number and mass.
5 0
3 years ago
A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?
pav-90 [236]
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
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4 0
3 years ago
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To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
Waves can transfer energy through
lina2011 [118]

Answer:

electromagnetic waves

Explanation:

"wave" is a common term for a number different ways in which energy is transferred

3 0
3 years ago
A circuit contains two light bulbs connected in parallel. What would happen to the brightness of each light bulb if two more lig
laila [671]

Answer;

C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.

Explanation;

-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.

-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source

6 0
3 years ago
Read 3 more answers
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