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tia_tia [17]
3 years ago
14

In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

gorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?
Physics
1 answer:
irinina [24]3 years ago
8 0

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_{1,2}= Mass of each object

v_{1,2} = Initial velocity of each object

v_f= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

m_2v_2 = (m_1+m_2)v_f

(0.42)(21) = (90+0.42)v_f

v_f = 0.0975m/s

Therefore the velocity right after catching the ball is 0.0975m/s

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Which best explains parallel forces?
Lyrx [107]
The answer is to this question is B
8 0
3 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

(d) Change in angle is given by

\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev

7 0
3 years ago
A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a
saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

  • Boyle's Law,   (P₁V₁ =P₂V₂)
  • Charles' Law  (V₁/T₁ =V₂/T₂)
  • Gay-Lussac's Law.  (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) \frac{5}{278} =\frac{P_2}{313}  \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm

7 0
3 years ago
A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg
dezoksy [38]

Answer:

For cast iron we have

h = 0.92 m

For copper

h = 1.05 m

For Lead

h = 1.23 m

For Zinc

h = 2.43 m

Explanation:

As we know that final speed of the block is calculated by work energy theorem

W_f + W_g = \frac{1}{2}mv^2

now we have

-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2

now we have

v^2 = 2gh - 2\mu_k g h cot\theta

v = \sqrt{2gh(1 - \mu_k cot\theta)}

For cast iron we have

4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}

h = 0.92 m

For copper

4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}

h = 1.05 m

For Lead

4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}

h = 1.23 m

For Zinc

4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}

h = 2.43 m

4 0
3 years ago
What is the answer to 895cm a Km
victus00 [196]

Answer:

if you're converting then the answer is 0.00895

Explanation:

895 centimetres converted into kilometres= 0.00895

8 0
2 years ago
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