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3 years ago

14

What is the difference between the velocity and the speed of an object?

1 answer:

timurjin [86]3 years ago

3 0

Answer:

Velocity has a direction associated with it, while speed has no specific direction.

Explanation:

Velocity is a vector, while speed is a scalar.

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When you stretch a spring 20 cm past its natural length, it exerts a force of 8

Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0

3 years ago

Read 2 more answers

Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw

aalyn [17]

Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons

4 0

3 years ago

A rock has a density of 2 g/mL and a mass of 17.5 g. What is the volume of the rock?

Lelu [443]

D=m/v then v=m/d=17.5/2=8.75ml

7 0

3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

bixtya [17]

Answer:

a. $P=18.61\ W$

b. $\eta_c=0.6976=69.76\%$

c. $\eta=0.4718=47.18\%$

Explanation:

Given:

temperature of the hotter reservoir, $T_H=1250\ K$

temperature of the colder reservoir, $T_C=378\ K$

heat absorbed by the engine, $Q_H=1.42\times 10^{5}\ J$

heat rejected to the cold reservoir, $Q_L=7.5\times 10^4\ J$

time duration of the energy transfer, $t=1\ hr=3600\ s$

<u>Now the work done by the engine:</u>

Using energy conservation,

$W=Q_H-Q_L$

$W=14.2\times 10^4-7.5\times 10^4$

$W=6.7\times 10^4\ J$

a.

<u>Hence the power output:</u>

$P=\frac{W}{t}$

$P=\frac{6.7\times 10^4}{3600}$

$P=18.61\ W$

b.

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{378}{1250}$

$\eta_c=0.6976=69.76\%$

c.

now actual efficiency:

$\eta=\frac{W}{Q_H}$

$\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}$

$\eta=0.4718=47.18\%$

6 0

4 years ago

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

IceJOKER [234]

Answer: $E=75\ N/m$

Explanation:

Given

First charge of $q_1=15\ nC$ is placed at $x=1.5\ m$

Second charge $q_2=-20\ nC$ is placed at $y=-2\ m$

Electric field is given by

$E=\frac{kq}{r^2}$

Electric field due to $q_1$ is away from it

$E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}$

$E_1=60\ N/m$

Electric field due to $q_2$

$E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}$

$E_2=45\ N/m$

Net electric field will be vector addition of two

$\vec{E_{net}}=\vec{E_1}+\vec{E_2}$

$\vec{E_{net}}=-60\hat{i}-45\hat{j}$

Magnitude of Electric field is

$E=\sqrt{60^2+45^2}$

$E=75\ N/m$

8 0

3 years ago