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3 years ago

6

9. Use the following information to answer parts a, b, c and d

1 answer:

N76 [4]3 years ago

7 0

Answer:

iittdhusludepeotlke is the only person who has ever seen a long distance friend from which he is the 3PM and 3AM and I

Explanation:

gt6e6ofkhithodzipfrhFfu0sa6eo5ostiatkstoss ragers the following 3AM is not the only one sentences you have to say about it is not an 2475232347 email from a substance abuse treatment center for a medical professional who is a substance addict or a physician in my life to be 367inches 4 to the end 6th

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When red-headed woodpeckers (melanerpes erythrocephalus) strike the trunk of a tree, they can experience an acceleration ten tim

kifflom [539]

98.1 m/s² × (1 in)/(0.0254 m) ≈ 3862 in/s²

8 0

3 years ago

An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. he holds one e

Rainbow [258]

Understanding the given:
85 kg mountain climber
6.50 m long rope
gravity = 10m/s2

If we want to identify the work done on this scenario
we get f = 85kg x 10m/s2 = 850 N
w = 850N x 6.5 m = 5525 J

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

4 0

3 years ago

The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 10.0 m

adell [148]

Answer:

5.565 V

Explanation:

Radius of coil of generator=r=0.14 m

Length of wire=l=10 m

Magnetic field,B=0.24 T

Angular speed, $\omega=34rad/s$

We have to find the peak emf of the generator.

$N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11$

$A=\pi r^2=\pi (0.14)^2=0.062m^2$

Peak(maximum) induced emf of generator= $E_{max}=NBA\omega$

Using the formula

$E_{max}=11\times 0.24\times 0.062\times 34$

$E_{max}=5.565 V$

3 0

3 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

3 years ago

Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

Harman [31]

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

$W=[ Fcos\theta]*S$ where $Fcos\theta$ is the component of the force along the direction of displacement.

$Hence\ W= FScos\theta$

$= F.S$

As per the question the power P is given as -

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

$=\frac{F.S}{\delta t}$

$= F.\frac{S}{\delta t}$

$= \ F.V$

Hence alternative definition of power P = F.V

8 0

3 years ago

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