All categories

3 years ago

Muscular Strength exercises focus on high____and low_

Physics

1 answer:

Advocard [28]3 years ago

4 0

Answer:

d d d d dd d d d d d dd d d d d dd d

You might be interested in

A planet moves forward because of momentum <br><br> True or False??

lorasvet [3.4K]

Answer:

true

Explanation:

8 0

3 years ago

How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or

Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

$r' = \frac{r}{2}$

So the new force will be

$F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F$

So, the force will quadruple.

4 0

3 years ago

Two sacks contain the same number of identical apples and are separated by a distance r. The two sacks exert a gravitational for

stepan [7]

Answer:

Explanation:

im sleppy

3 0

3 years ago

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

5 0

3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

5 0

3 years ago

Muscular Strength exercises focus on high______and low___

Muscular Strength exercises focus on high____and low_