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svet-max [94.6K]

1 year ago

13

g) If two cars A and B are moving with velocity 60 km/hr and 80 km/hr respectively in the same direction. What will be the relat

ive velocity of B with respect to A? (20 kmhr)

1 answer:

tester [92]1 year ago

4 0

Explanation:

Here,

Velocity of A=60km/hr

Velocity of B=80km/hr

Now,

Relative velocity of B with respect to A in the same direction=Velocity of B - Velocity of A

=80km/hr-60km/hr

=20km/hr

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Pitch describes how high or low a sound is. The pitch of a sound is most dependent upon the of the sound wave.

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A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away

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Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d) Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it. Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball. Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.

Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

x = x(0) + V_i,x*t_c

D = 0 + V_i,x*t_c

V_o,x + v_r = D/t_c

V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

V_o = sqrt ( V_o,x^2 + V_o,y^2)

V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

- The angle with which it should be thrown can be evaluated by trigonometric relation:

tan(Q) = ( V_o,y / V_o,x )

tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

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2 years ago

Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum

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Answer:

$v_{top} = 61.96 \frac{mi}{h}$

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To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

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$v_{top} = 27.7 \frac{m}{s}$

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

$1609.34 \ m = 1 \ mi$

Now, we can divide by 1609.34 meters on both sides:

$\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}$

The left sides equals 1, so

$1 = \frac{1 \ mi}{ 1609.34 \ m}$

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 0.0172120 \frac{mi}{s}$

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

$1 \ h = 3600 \ s$

as the seconds are dividing in the velocity, we know divide by 1 hour.

$\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}$

$1 = \frac{3600 \ s}{ 1 \ h}$

and know we just multiply our top speed by this conversion factor

$v_{top} = 0.0172120 \frac{mi}{s} \frac{3600 \ s}{ 1 \ h}$

$v_{top} = 61.96 \frac{mi}{h}$

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