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3 years ago

What is the impulse of a 2 kg object that starts at 10 m/s and comes to a stop over 0.4 seconds?

Physics

1 answer:

gizmo_the_mogwai [7]3 years ago

8 0

Answer: Impulse = 20 Ns

Explanation:

Impulse is the product of force and time

Also impulse = momentum

Where momentum is the product of mass and velocity.

Given that

M = 2kg

V = 10 m/s

Impulse = MV = 2 × 10 = 20 Ns

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How can you find directions using satellite orbiting?

quester [9]

Answer:

Satellites may move north to south, or south to north, or west to east, but never from east to west. When satellites are launched, they always head eastward to take advantage of the Earth's rotation, going more than 1,000 miles per hour near the equator. This saves a lot of fuel.

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3 years ago

Which statement best describes the adiabatic process? A.the temperature remains constant B.the temperature increases at a consta

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3 years ago

A cylindrical solenoid (radius r=0.6 m, turns N=600, length l=0.5 m) carries 15 A of current. What is the inductance L of the co

Anuta_ua [19.1K]

Answer:

L = 1.023 H

Explanation:

given,

radius of the cylindrical solenoid = r = 0.6 m

Number of turns = N = 600

Length = l = 0.5 m

Current in the cylindrical solenoid = 15 A

Inductance in the coil = ?

using formula

$L = \dfrac{\mu_0\ N^2\ A}{l}$

$L = \dfrac{4 \pi \times 10^{-7}\times 600^2\times \pi r^2}{0.5}$

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L = 1.023 H

the inductance L of the coil is = 1.023 H

8 0

3 years ago

Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp

yuradex [85]

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

7 0

3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

$\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz$

8 0

3 years ago