All categories

3 years ago

What question could you ask about kinetic energy which will include the variables that affect it?

Physics

2 answers:

Gemiola [76]3 years ago

4 0

Answer:

<u>Kinetic Energy (K.E):</u>

The K.E is acquired or possessed by the body due to its virtue of motion. As the K.E is based upon the mass of the body, m which remains the same for each object, while the K.E also depends upon the square of velocity,v of the same body(while it is in motion), which is a variable.

K.E=1/2 m×v².

Explanation:

<u>K.E and velocity,v:</u>

The K.E acquired by the body depends upon the constant value of mass,m and the variable component of velocity,v. As the velocity of the body changes along time change,Δt.

Juli2301 [7.4K]3 years ago

3 0

KE= (1/2) mv²

So, the variables we need to include in our question would be a varable for a mass(m) of an object at some velocity(v).

My Answer:

(This is just an example question, yours can be different)
What is the Kinetic Energy experienced by an bouncy ball rolling at 7m/s (that's your velocity) across a frictionless surface that has a mass(m) of 10 grams?

You might be interested in

The property of matter that resists changes in motion is

vredina [299]

Answer:

Inertia

Explanation:

Inertia is the property that any physical object has of remaining in its state of relative motion. Therefore, it is the resistance that opposes matter to modify its state of motion, which includes changes in speed or changes in the direction of movement.

6 0

3 years ago

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

B1 at 20km/h

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago

How to find the magnitude and direction of a resultant velocity?

mixas84 [53]

Find the horizontal components vcos30 ...one goes right and one goes left so they cancel each other.
Find vertical components vsin30.....there are two of them.... so 2vcos30....hey presto... resultant velocity = 2vCos30

5 0

2 years ago

A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *

ad-work [718]

Answer:

D) 15 m/s

Explanation:

v = u + at

v = 10 m/s + 1 m/s²(5 s)

v = 15 m/s

3 0

2 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago