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goblinko [34]
3 years ago
12

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

on between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Physics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

mv_{1i}+Mv_{2i}=(m+M)v  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

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Can someone please help me ​
Tamiku [17]

<em>F</em> = 153 N

\theta = 11.3°

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components F_{x} and F_{y}:

F_{x} = 350 N - 200 N = 150 N

F_{y} = 180 N - 150 N = 30 N

The magnitude of the resultant force <em>F</em> is given by

F = \sqrt{F_{x}^{2}+F_{y}^{2}}

\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}

\:\:\:\:\:\:=153\:N

To find the direction \theta, we use

\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2

or

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4 0
3 years ago
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 when the han
slava [35]

There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer:

t=3.37s

Explanation:

Given Data

As we have taken hand at origin and positive upward

So given data are

y_{i}=0m\\y_{f}=-1.80m\\v_{i}=16.0m/s\\a=g=9.8m/s^{2}

To find

time taken by the ball before it hits the ground

Solution

By using the common kinematic equation

y_{f}=y_{i}+v_{i}t+0.5at^{2}

Put the given values and find for t

So

-1.80=0+16.0t+(0.5*(-9.8)t^{2} )\\-1.80=16.0t-4.9t^{2}\\ 4.9t^{2}-16.0t-1.80=0

Apply quadratic formula to solve for t

t=\frac{-(-16.0)+\sqrt{(-16)^{2}+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s

4 0
3 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.4Vwhen operating in a circuit with two lightbulbs, each rated at
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Answer:

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Explanation:

P = Power of bulb = 4 W

V = Voltage = 12 V

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow R=\dfrac{V^2}{P}\\\Rightarrow R=\dfrac{12^2}{4}\\\Rightarrow R=36\ \Omega

The connections are in Parallel

Equivalent resistance of these two bulb

\dfrac{36}{2}=18\ \Omega

Current from the source

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The voltage drop is

12-11.4=0.6\ V

Voltage drop is given by

\Delta V=R_iI\\\Rightarrow R_i=\dfrac{\Delta V}{I}\\\Rightarrow R_i=\dfrac{0.6}{0.63}\\\Rightarrow R_i=0.952380952381\ A

The internal resistance is 0.95 A

3 0
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