1250kgm²/s is the motional kinetic energy of a 25kg object moving at a speed of 10m/s
Kinetic energy of an object is defined as the energy which is possessed when that is in motion. It is the energy of the kinetic mass of an object. Kinetic energy is never negative and is a scalar quantity. That is, it shows only size, not orientation.
Given to us
Mass of the object, m=25kg
Velocity of the object, v=10m/s
K.E=1/2x25x10²
=1250
Kinetic energy is directly proportional to the mass and velocity squared (K.E.) of an object. =1/2xMxV². If the mass is in kilograms and the velocity is in meters/second, then the kinetic energy is in kilograms - meters squared/second.
Learn ore about Kinetic energy here brainly.com/question/25959744
#SPJ9
Answer:
B. Cant stop things from going wrong.
Explanation:
To me it's the only reasonable answer...
Complete Question
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is 
Compute U if the bulb remains on for 5h
Answer:
The value is 
Explanation:
From the question we are told that
The power rating of the bulb is
The resistance is 
The voltage is ![V = V_o sin [2 \pi ft]](https://tex.z-dn.net/?f=V%20%20%3D%20%20V_o%20%20sin%20%5B2%20%5Cpi%20ft%5D)
The energy expanded is 
The voltage 
The frequency is 
The time considered is 
Generally power is mathematically represented as

=> ![P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%28%20110%20%20sin%20%5B2%20%5Cpi%20%2A%2060t%5D%29%5E2%7D%7B%20144%7D)
=> ![P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}](https://tex.z-dn.net/?f=P%20%3D%20%20%5Cfrac%7B%20110%5E2%20%5B%20sin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D)
So
![U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cint%5Climits%5ET_0%20%7B%20%5Cfrac%7B%20110%5E2%2A%20%20%5Bsin%20%5B120%20%5Cpi%20t%5D%29%5E2%7D%7B%20144%7D%7D%20%5C%2C%20dt)
=> ![U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt](https://tex.z-dn.net/?f=U%20%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5Cint%5Climits%5ET_0%20%7B%20%28%20%20%20sin%5E2%20%5B120%20%5Cpi%20t%5D%7D%20%5C%2C%20dt)
=> 
=> 
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%20T%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
=> ![U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7Bt%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20t%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D%5Cleft%20%20%7C%2018000%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]](https://tex.z-dn.net/?f=U%20%3D%20%20%5Cfrac%7B110%5E2%7D%7B144%7D%20%5B%5Cfrac%7B18000%7D%7B2%7D%20%20-%20%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Cfrac%7Bsin%28240%20%5Cpi%20%2818000%29%29%7D%7B240%20%5Cpi%7D%20%5D%20%5D)
=> 
Explanation:
Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).
Answer:
t = 0.319 s
Explanation:
With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by
v = √T/λ
Linear density is
λ = m / L
λ = 4/20
λ = 0.2 kg / m
The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg
T = W = mg
T = 80 9.8
T = 784 N
The pulse rate is
v = √(784 / 0.2)
v = 62.6 m / s
The time it takes to reach the hook can be searched with kinematics
v = x / t
t = x / v
t = 20 / 62.6
t = 0.319 s