A referee will toss up the ball between to opponents.what is this called

Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and

tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

$M = 1.0 kg$

$r = 0.5 m$

$m = 10 g$

$I = 0.280 kg.m^2$

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

$I = I_{rim} + n * I_{spoke}$

Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

4 years ago

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

iragen [17]

For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0

2 years ago

A 63 kg kg person starts traveling from rest down a waterslide 6.0 mm above the ground. At the bottom of the waterslide, it then

gladu [14]

Answer:

change waterslide according to question. and you are good to go. check photo for solve

5 0

3 years ago

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

$A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\$

since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

6 0

4 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.