On an acceleration-time graph, what are the units for acceleration

Suppose a skydiver (mass =100kg) is falling towards the earth. When the skydiver is 80 m above the earth he is moving at 60 m/s

goblinko [34]

Answer:

The total mechanical energy of the skydiver is, E = 96402.6 J

Explanation:

Given data,

The mass of the skydiver, m = 100 kg

The speed of the skydiver at 80 m height, v = 60 m/s

The initial velocity of the skydiver, u = 0

Using the III equations of motion,

v² = u² + 2gs

s = v²/2g

Substituting the given values,

s = ½ 60²/ 9.8

= 18.37 m

Hence the initial total distance of the skydiver from the ground initially,

h = s + d

= 18.37 + 80

= 98.37 m

Since the total mechanical energy of a system is conserved, the total mechanical energy of the skydiver at height 'h' is equal to the total mechanical energy at height 'd'.

E = P.E + K.E

= mgh + ½ mu²

= 100 x 9.8 x 98.37 ( ∵ u = 0)

= 96402.6 J

Hence, the total mechanical energy of the skydiver is, E = 96402.6 J

5 0

3 years ago

a tennis ball is hit upward with a tennis racket an initial velocity of 12 m/s. What will the ball's speed be when it returns to

zaharov [31]

Answer: 12 m/s

Explanation:

If we ignore air resistance, gravity alone will reduce the upward velocity to zero at the top of the flight. As gravity is a conservative force, it will return exactly the same amount of energy to the tennis ball when it returns to the original elevation.

7 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago

How do you solve #3 and #5?

julia-pushkina [17]

Answer:

your answers are correct i have done this many times

8 0

3 years ago

Suppose you want to represent the velocity of a BALL with respect to a particular frame of reference from aboard a moving TRAIN.

tatyana61 [14]

Answer:

Explanation:

We would most likely write the velocity of the ball as follow :

V(b<em>all</em> with respect to t<em>rain)</em> = Vbt

4 0

3 years ago