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Deffense [45]
3 years ago
7

A bird, accelerating from rest at a constant rate, experiences a displacement of 28 m in 11 s. what is the final velocity after

11 s?
Physics
1 answer:
frutty [35]3 years ago
4 0
Let's see what variables we've got first. Hmmm. We have:

Displacement, d = 28 m
Time taken, t = 11 s
Initial velocity, u = 0 m/s (at rest)

And now we need to find the final velocity, v. Among the 4 (or 5) equations of motions, there's no equation that will let us simply plug in the values and give an answer sigh. But fear not! We'll do it in steps.

I'm going to pick one of the motion equation to find more information:

d = ut +  \frac{1}{2} a {t}^{2}

I know everything except for a in this one, so I I'll use this! After plugging in values, I get a = 0.4628 m/s^2.

Now I'm going to use another motion equation that has v in it because that needs to be solved!

v = u + at

Now I know everything except dial velocity v. Nice!

v = 0 + (0.4628)(11)
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Answer:

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3 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
4 years ago
The potential energy of a 40-kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
nasty-shy [4]
Pe = mgh.
14000 J = (40kg)(10m/s^2)(h)
h = 35 meters
7 0
3 years ago
Read 2 more answers
What is the main way in which heat transfer occurs in liquids and gases?
Taya2010 [7]
The answer is convection
7 0
3 years ago
A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of m
tatiyna

Answer:

1. e m fmax = 0.00598 Volt

2. Imax = 0.000854 Amp

Explanation:

1. Find the maximum induced emf.

e m fmax =

Given that e m fmax = N*A*B*w

N = 1

A = 2 cm^2 = 0.0002 m^2

f = 143 rotation per minute = 143/min

f = (143/min) * (1 min/60 sec) = 2.38/sec

w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec

B = 2T

e m fmax = N*A*B*w

e m fmax = 1 * 0.0002 * 2 * 14.95

e m fmax = 0.00598 Volt.

2. Find the maximum current through the bulb.

Imax = e m fmax / R

Where R is the total resistance in the circuit is 7 Ω.

Imax = 0.00598/7 = 0.000854 Amp.

Imax = 0.000854 Amp

6 0
3 years ago
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