If do 50 J of Work in 20 seconds, what is my Power ?

A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill

Stolb23 [73]

Answer:

a.) the speed at the bottom is greater for the steeperhill

Explanation:

since the energy at the bottom of the steeper hilis greater

$mgh =\frac{1}{mv^2}$

As we can see from above that v is higher when h ishigher.

8 0

3 years ago

The image to the left shows Earth’s major plates. A geologist is studying a plate boundary indicated by the arrow. Fill in the b

nexus9112 [7]

Answer:

The Eurasian Plate

Explanation:

The Eurasian plate is one of the most extended on Earth, crossing all of Asia and Europe. The Eurasian plate is between the North American and the African Plates on the north and west sides. The Eurasian plate crushed up above the Indian plate. The Tibetan plateau and the Himalayan mountain range formed due to the crush between the Eurasian Plate and Indian Plate, which started 50 million years ago.

3 0

3 years ago

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul

Gnom [1K]

Answer:

$t=2.5\times 10^{-14}\ s$

Explanation:

We know that charge on electron

$q=1.6\times 10^{-19}\ C$

r= 2 nm

We know that force between two charge given

$F=K\dfrac{Q_1Q_2}{r^2}$

Now by putting the value

$F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}$

$F=5.67\times 10^{-11}\ N$

We know that mass of electron

The mass of electron

$m=9.1\times 10^{-31}\ kg$

F= m a

a= Acceleration of electron

a= F/m

$a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2$

$a=6.2\times 10^{19} m/s^2$

$S=ut+\dfrac{1}{2}at^2$

initial velocity given that zero ,u=0

$20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2$

$t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}$

$t=2.5\times 10^{-14}\ s$

3 0

3 years ago

What is the best free energy source?<br> Nuclear<br> Solar<br> Natural Gas

Free_Kalibri [48]

Solar it is the cheapest and widely used energy source

4 0

3 years ago

Velocity vector and acceleration vector in a uniform circular motion are related as.

mr_godi [17]

They are related as $\bold{\underline {v}\,.\,\underline a }= \bold{0}$

In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
In mathematics, 2 vectors ( $\underline p$ , $\underline q$ ) that are perpendicular to each other have a quality that their dot product ( $\underline p\,.\, \underline q$ ) equal to zero vector ( $\bold 0$ ) which is written as $\undeline p\,.\, \underline q = \bold 0$ .
This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner $\underline v\,.\, \underline a =\bold 0$ .

#SPJ4

8 0

1 year ago