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3 years ago

H.e.y w.h.o w.a.nt.s t.o t.a.l.k

Physics

2 answers:

butalik [34]3 years ago

6 0

Answer:

nope

Explanation:

ivanzaharov [21]3 years ago

4 0

Answer: Whats up!!!

Explanation:

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Which has more mass - a 5-kg bag of feathers or a 5-kg<br> cannon ball?

Semenov [28]

Answer:equal mass

Explanation:the mass of both is 5kg

7 0

2 years ago

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

muminat

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.

KE = 1/2*m*v^2

= 1/2*0.005*320^2

= 256 Joules.

Divide this by the distance over which this energy was received and you have the force that provided that energy.

= 256/0.820 = 312.195 Newtons

Rounded off, this is 312 N

3 0

3 years ago

A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0

Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

$M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\$

If we carry out summation of Torques on the pulley we will have to,

$F_2*d-F_1*d = I \alpha$

Where,

I = Inertia moment

$\alpha =$ Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

$I = \frac{1}{2} mr^2$

Replacing this equations we have know that

$(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})$

$F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}$

$F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1$

$F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}$

Replacing our values we have that

$F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}$

$F_1 = 0.974 N$

Therefore the tension in the string between the pulley and the cart is 0.974 N

6 0

4 years ago

Define "increased productivity" in terms of the number of tasks and the amount of time.

vovikov84 [41]

Answer:

The greater the amount of output for a given unit of input, the higher the overall productivity. Businesses generally aim to improve productivity over time to maintain competitiveness and increase the business's profitability. Individuals are familiar with the idea of productivity in their own lives.

6 0

3 years ago

A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper

vodka [1.7K]

Answer:

t = 402 years

Explanation:

To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:

$I=nqv_dA$ (1)

n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3

q: charge of the electron = 1.6*10^-19 C

vd: drift velocity of electron in the metal = ?

A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2

I: current in the wire = 1110 A

You solve the equation (1) for vd:

$v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s$

Next, you calculate the time by using the information about the length of the line transmission:

$x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years$

hence, the electrons will take aproximately 402 years in crossing the line of transmission

6 0

3 years ago