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2 years ago

6

Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b

all, he and the ball move on the skateboard. How fast do Jocko and the ball move after he catches the ball?

1 answer:

Mekhanik [1.2K]2 years ago

7 0

Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

Mass if Jocko, m = 60 kg

Mass of the ball, m' = 20 kg

Speed of the ball, v = 3 m/s

Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

$m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s$

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

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A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)

</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>

8 0

2 years ago

Read 2 more answers

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is $\tau = 48t \r k$

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration

ollegr [7]

Answer:

<h3>The answer is 4.29 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken acting on it can be found by using the formula

$acceleration = \frac{velocity}{time} \\$

From the question

velocity = 30 m/s

time = 7 s

We have

$acceleration = \frac{30}{7} \\ = 4.285714...$

We have the final answer as

<h3>4.29 m/s²</h3>

Hope this helps you

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2 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position is given by

$\Delta h=L(1-\cos \theta )$

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

$mgL(1-\cos \theta )=\frac{mv^2}{2}$

$v=\sqrt{2gL(1-\cos \theta )}$

$v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}$

$v=\sqrt{4.021}$

$v=2 m/s$

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2 years ago

The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe

Novosadov [1.4K]

Answer:

a) a= 8.33 m/s², T = 12.495 N , b) a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

fr₁ + fr₂ = ma

The equation for the force of friction is

fr = μ N

Y Axis

N - W₁ –W₂ = 0

N = W₁ + W₂

N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

N = 2 m g

We replace

μ₁ 2mg + μ₂ mg = m a

a = (2μ₁ + μ₂) g

a = (2 0.30 + 0.25) 9.8

a= 8.33 m/s²

Let's look for cable tension with beam 2

T = m₂ a

T = 1500 8.33

T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

fr = m₂ a₂

N- w₂ = 0

N = W₂ = mg

μ₂ mg = m a₂

a = μ₂ g

a = 0.25 9.8

a = 2.45 m / s²

4 0

2 years ago