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3 years ago

a 1 gram spiders sits on a platform rotating at 78 rpm. the spider is 15 cm from the centre disk. find the speed of the spider

1 answer:

8 0

The spider is traveling in a circle with radius = 15cm

The circumference of any circle = <em>2 pi (radius)</em>
The circumference of the spider's path = 2 pi (15 cm) = 30 pi cm

The spider completes a trip around this path 78 times per minute.
Its speed, relative to you, is

   (78) x (30 pi) cm/min =

   2,340 pi cm/min = 7,351.33 cm/min =

   <em> 73.5133 meter/min =</em>

   <em>4.411 km/hr =</em>

   <em>2.74 miles/hour

</em>(After the last appearance of pi,
all numbers are rounded.)<em>

</em>

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blondinia [14]

The answer is “ Mass “

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3 years ago

Holes are being steadily injected into a region of n-type silicon (connected to other devices, the details of which are not impo

Tems11 [23]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $J_n = -0.864 \ A/cm^2$

Explanation:

Generally for an n-type semiconductor the current density is mathematically represented as

$J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

Here $p_{n_o}$ is mathematically represented as

$p_{n_o} = \frac{n_i^2}{n_d}$

=> $p_{n_o} = \frac{(1.5*10^{10})^2}{10^{16}}$

=> $p_{n_o} = 2.25 *10^{4} \ cm^{-3}$

$d p_{n_o} = P_n0 - p_{n_o}$

From the diagram $P_n0 = 10^8 * p_{n_o}$

=> $P_n0 = 10^8 * (2.25 *10^{4} )$

$d p_{n_o} = 10^8 * (2.25 *10^{4} ) - 2.25 *10^{4}$

$d p_{n_o} = 2.25 *10^{12} cm^{-3}$

So from $J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

substitute

$1.60 *10^{-19} \ C$ for q and $w_2 =50 nm = 50*10^{-9} m = 5*10^{-6} cm$

and from the diagram $w_1 =0 \ cm$

$J_n = -1.60 *10^{-19} *12 * \frac{2.25 *10^{12} }{ 5*10^{-6} - 0 }$

$J_n = -0.864 \ A/cm^2$

6 0

3 years ago

Click on the reset button, and stack one 50kg. crate on top of the other, so that the total mass is 100kg. The Friction should b

Helen [10]

Answer:

490.5 N

Explanation:

Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction

F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.

F=kmg=0.5*100*9.81=490.5 N

6 0

3 years ago

17 points

const2013 [10]

The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>

MV = mv................. Equation 1

<h3>Where:</h3>

M = mass of the cannon
m = mass of the cannonball
V = velocity of the cannon
v = velocity of the cannonball

Make v the subject of the equation.

v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>

M = 500 kg
V = 3 m/s
m = 10 kg.

Substitute these values into equation 2.

v = (500×3)/10
v = 150 m/s.

Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

2 years ago

A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed int

Stells [14]

Answer:

A. The hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand = 1025 N

Explanation:

A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.

Workdone = force × distance

Minimum force required = 870 N;

Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M

Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J

This work done = Kinetic energy of the hand

Kinetic energy = mv²/2 ; where m is mass and v is velocity

Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J

v² = 2 KE/m

v = √2KE/m

v = √(2 × 12.18/0.50)

v = 6.98 m/s

Therefore, the hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand

This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration

From the equation of motion, a = v² - u²/2s

At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m

a = 0² - 6.98²/ 2 × 0.012

a = -2030 m/s²

Force = 2030 m/s² × 0. 50 kg = 1015 N

Therefore, Average force on the hand = 1025 N

3 0

3 years ago