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3 years ago

Snowboarder Jump—Energy and Momentum

Physics

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

THE ANSWER TERMS ARE DEFINED BLOW:-

Explanation:

MOMENTUM- IT IS THE ABILITY TO INCREASE OR DEVELOP CONSTANT FORCE.

KINETIC ENERGY:- IT IS THE ENERGY THAT A PRTICLE POSSES WHEN IT IS ACTUALLY IN MOTION.

POTENTIAL ENERGY:- IT IS THE ENERGY THAT A PARTICLE POSSES WHEN IT ACTUALLY IS IN RESTING STATE.

IN THIS ACIVITY THE SNOWBOARDER IS IN THE MOTION STATE THEREFORE HE POSSES KINETIC ENERGY AND TO MAINTAIN THAT KINEITC ENERG FOR A PERIOD OF TIME,MOMENTUM PLAYS IT'S ROLE.

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A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes

Marat540 [252]

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

3 0

2 years ago

Một cần trục có trọng lượng Q = 50 kN cẩu vật nặng có trọng lượng P = 10 kN

lesantik [10]

3373727227717177

Explanation:

yun hehew

8 0

2 years ago

Read 2 more answers

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0

3 years ago

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

solong [7]

Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s

3 0

3 years ago

A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

4 0

3 years ago