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3 years ago

Why do quarterbacks throw the football with significant spin about its long axis?

Physics

1 answer:

Lera25 [3.4K]3 years ago

7 0

(a) the principle of aerodynamic convergence

(b) the centripetal force

(d) Conservation of kinetic energy

(e) None of these

Conservation of angular momentum

Answer: Option C.

<u>Explanation:</u>

The law of conservation of angular momentum expresses that when no outer torque follows up on an article, no difference in precise force will happen.

The law of conservation of angular momentum expresses that the angular energy of a body that is the result of its snapshot of latency about the hub of revolution and its rakish speed about a similar pivot, can't change except if an outside torque follows up on the framework.

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A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

zhannawk [14.2K]

Answer:

$y = 17.04 ft$

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

$y = 180 ft$

horizontal distance of the tower from mid point is given as

$x = \frac{1300}{2} = 650 ft$

now from the equation of parabola we have

$y = k x^2$

$180 = k(650^2)$

$k = 4.26 \times 10^{-4}$

now we have

$y = (4.26 \times 10^{-4})x^2$

now we need to find the height at distance of 200 ft from center

so we have

$y = (4.26 \times 10^{-4})(200^2)$

$y = 17.04 ft$

8 0

3 years ago

Anna pushes a box with a force of 8.00 newtons. She generates a power of 3.00 watts. How much time does it take for Anna to move

QveST [7]

Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:

P = E / t

where E = Force x distance

P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s

8 0

3 years ago

Read 2 more answers

The format specifier ________ is a placeholder for an int value. %d %n %int %s

Grace [21]

%d is a format specifier that is a placeholder for an int value. It tells the compiler that we want to print an integer value that is present in variable a. In this way there are several format specifiers in c.

8 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$