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avanturin [10]
2 years ago
14

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t

he pitcher's hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?
Group of answer choices
A) 10.5 rad/s
B) 0.0952 rad/s
C) 0.00857 rad/s
D) 117 rad/s
Physics
1 answer:
stiv31 [10]2 years ago
3 0

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s

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Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp
Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

           v² = v₀² + 2 a x

           0 = v₀² + 2 a x

           a = - v₀² / 2 x

           

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

         a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

           

       a/g = - (2v₀) 2 / 2xg

       a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

        a/g = (- v₀² / 2g) 2/x

         

        a =2  (-v₀² / 2x₀g)  g

       

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

        a / g = - 20² / (2 0.05)

        a / g = - 4000 / g

        a / g = - 4000 / 9.8 = 408

        a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

6 0
2 years ago
The battery capacity of a lithium ion battery in a digital music player is 750 mA-h. The manufacturer claims that the player can
Oksi-84 [34.3K]

Answer:

The number of electrons is 6.3\times10^{21}\ electrons

(D) is correct option.

Explanation:

Given that,

Battery capacity = 750 mA-h

Time t= 8 hours

Time t'=3 hours

We need to calculate the battery capacity

Battery\ capacity=750\times10^{-3}\times3600

Battery\ capacity=2700\ A-s

We need to calculate the number of electrons in 1 C Li

Using formula for number of electron

n=\dfrac{1}{e}

n=\dfrac{1}{1.6\times10^{-19}}

n=6.25\times10^{18}\ electrons

We need to calculate the number of electron in 2700 C

2700\ C=2700\times6.25\times10^{18}=1.68\times10^{22}\ electrons

The total number of electrons battery can deliver in 8 hours

n=1.68\times10^{22}\ electrons

We need to calculate the number of electron in 3 hours

Using formula of number of electrons

n=\dfrac{n\times t'}{t}

Put the value into the formula

n=\dfrac{1.68\times10^{22}\times3}{8}

n=6.3\times10^{21}\ electrons

Hence, The number of electrons is 6.3\times10^{21}\ electrons

8 0
2 years ago
Read 2 more answers
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
2 years ago
Hydrogen (H-1), deuterium (H-2), and tritium (H-3) are the three isotopes of hydrogen. What are the values of a, b, and c in the
asambeis [7]

I believe it is a=1 b=2 c=3

plato students

6 0
2 years ago
Read 2 more answers
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