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Anika [276]
3 years ago
8

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

ng?
Physics
1 answer:
vovangra [49]3 years ago
4 0

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

        Momentum = mass x velocity

 Since the speed of the ball is unknown:

           Velocity  =  \frac{momentum }{mass}

                           = \frac{0.78}{0.15}

                           = 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

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Answer:

2587.2 J.

Explanation:

From potential energy,

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W = mgh............... Equation 1

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But,

m = m'd............... Equation 2

Where m' = density of the chain, d = length of the chain.

Substitute equation 2 into equation 1

W = m'dgh................ Equation 3

Given: m' = 2 kg/m, d = 12 m, h = 11 m, g = 9.8 m/s²

Substitute into equation 3

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4 years ago
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
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Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

v_f = v_i + at

v_f = v_o - gt

Part d)

distance from helicopter

d = \frac{1}{2}gt^2

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3 years ago
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The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
ivolga24 [154]

Answer: 0.10233nm

Explanation:

The mean free path \lambda   of an atom is given by the following formula:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is the Universal gas constant

T=0\°C=273.115K is the absolute standard temperature

d is the diameter of the helium atom

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

Knowing this, let's find d from (1), in order to find the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Finally:

r=0.10233nm This is the radius of the helium atom in nanometers.

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