Which contains the particles that are vibrating fastest?

If a ball is given a push so that it has an initial velocity of 5 m/s down a certain inclined plane, then the distance it has ro

iren [92.7K]

Answer:

v= 13 m/s

Explanation:

Velocity is defined as the derivative of displacement with respect to time

v= ds/dt

Known data

s(t) = 5t + 2t² : distance that the ball has rolled after t seconds

vi= 5 m/s : initial velocity

t= 2 s

Problem develoment

s(t) = 5t + 2t²

v= ds/dt= 5 + 4t : velocity of the ball in function of the time

We replace t =2 s in the equation of velocity

v= 5 + 4(2)

v= 13 m/s : velocity after 2 seconds

4 0

3 years ago

Answer key says b but i think it is false please help me

klasskru [66]

Energy in a spring:

E = 0.5 * k * x²

k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m

E = 0.5 * 800 * 0.05² = 1

5 0

2 years ago

A ball is dropped from a cliff. determine how far the ball fell after 7.5 seconds

grin007 [14]

Answer:

The ball fell 275.625 meters after 7.5 seconds

Explanation:

If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are

$V_f=gt$

$\displaystyle y=\frac{gt^2}{2}$

Where $V_f$ is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object

The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula

$\displaystyle y=\frac{gt^2}{2}$

$\displaystyle y=\frac{(9.8)(7.5)^2}{2}$

$Y=275.625\ m$

5 0

3 years ago

I need help ASAP please :)

rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0

2 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: