What voltage (potential difference)must be applied to a 26 1

By approximately how much does your gravitational potential energy change when you jump

Svet_ta [14]

Potential Energy = height*mass*gravity
the height is factually the height of the center of mass, which will be varied the same as the lowest of your feet. So when you hurdle, the height that you jump can be used to distinguish the alteration in height of your center of mass.In other words, it is inclined on the mass of the human, be contingent on how extraordinary the person can jump. Supposing a man is 80 kg and he can jump such that his center of mass rises 0.3 m in Earth's standard gravitational field. This sums to approximately 240 Joules of increase in GPE.

4 0

3 years ago

A 22.3 N crate is resting on the floor.

vitfil [10]

weight of the crate is given as W = 22.3 N

now as per force balance condition

Net weight of the object is counter balanced by normal force on the object by the surface

so here we will have

$W = F_n$

$F_n = 22.3 N$

So the normal force acting on the object will be same as as the weight of the object which is 22.3 N

8 0

3 years ago

A disc-shaped grindstone with mass 50 kg and diameter 0.52m rotates on frictionless bearings at 850 rev/min. An ax is pushed aga

gladu [14]

Answer:

0.48

Explanation:

m = mass of disc shaped grindstone = 50 kg

d = diameter of the grindstone = 0.52 m

r = radius of the grindstone = (0.5) d = (0.5) (0.52) = 0.26 m

w₀ = initial angular speed = 850 rev/min = 850 (0.10472) rad/s = 88.97 rad/s

t = time taken to stop = 7.5 sec

w = final angular speed 0 rad/s

Using the equation

w = w₀ + α t

0 = 88.97 + α (7.5)

α = - 11.86 rad/s²

F = normal force on the disc by the ax = 160 N

μ = Coefficient of friction

f = frictional force

frictional force is given as

f = μ F

f = 160 μ

Moment of inertia of grindstone is given as

I = (0.5) m r² = (0.5) (50) (0.26)² = 1.69 kgm²

Torque equation is given as

r f = I |α|

(0.26) (160 μ) = (1.69) (11.86)

μ = 0.48

4 0

3 years ago

What is the source of energy across the entire spectrum

aleksklad [387]

The answer to your question is Sun.

8 0

3 years ago

In a cathode-ray tube, a beam of electrons is projected horizontally with a speed of 1.0x109 cm/s into a region between a pair o

jekas [21]

(a) $2\cdot 10^{-9} s$

The time it takes for the electron to pass through the plates depends only on its horizontal motion,

The horizontal motion is a uniform motion, with constant velocity

$v_x = 1.0\cdot 10^9 cm/s$

The total distance travelled by the electron to pass through the plates is

d = 2.0 cm

Using the equation for uniform motion,

$v_x = \frac{d}{t}$

where t is the time taken. Solving for t,

$t=\frac{d}{v_x}=\frac{2.0}{1.0\cdot 10^9}=2\cdot 10^{-9} s$

(b) -0.2 cm

The vertical motion of the electron is a uniform accelerated motion, so the vertical displacement is given by

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

where

$u_y$ is the initial vertical velocity

$a=-1.0\cdot 10^{17} cm/s^2$ is the acceleration (negative because it is downward)

We can write that

$u_y=0$

Since the electron initially travels horizontally, and so if we substitute the time we found in part (a), we can find the vertical displacement after the electron has passed through the plates:

$y=\frac{1}{2}at^2=\frac{1}{2}(-1.0\cdot 10^{17})(2\cdot 10^{-9})^2=-0.2 cm$