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Vedmedyk [2.9K]
3 years ago
15

6. A man is riding a bike with an acceleration of 5.0 meters per second and a mass of the bike and him combined is 20 kg. What f

orce is produced?
Physics
1 answer:
KonstantinChe [14]3 years ago
3 0

Newton's 2nd law of motion:

                         Force = (mass) x (acceleration)

For this man
on his bike:        Force = (20 kg) x (5 m/s²)

                                    =    100 newtons .

This is NOT the "force produced".

This is the force NEEDED in order to accelerate THAT man
and his bike at THAT rate.  If you don't have something that
can apply THAT force, then you can't create this situation,
and he'll just have to carry on, in the same direction and at
the same speed, without turning, slowing down, or speeding up.

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A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor
OLEGan [10]

Answer:

(e)6.835\times 10^{-7}F

Explanation:

At resonance we know that X_l=X_C

That is \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

\omega =\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So 4511=\frac{1}{2\pi \sqrt{LC}}

LC=\frac{1}{4\pi ^2\times 4511^2}

LC=1.244\times 10^{-9}

C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F

So option e is the correct answer

3 0
3 years ago
A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire
Debora [2.8K]

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

P = F.v

P = 4 W

v = 4 m/s

now we will have

4 = F*4

F = 1N

So external force required will be 1 N

PART B)

now in order to find magnetic field strength we can say

P = \frac{v^2B^2L^2}{R}

here we know that induced EMF in the wire is E = vBL

so power due to induced magnetic field is given by

P = \frac{E^2}{R}

4 = \frac{4^2*B^2*0.10^2}{0.20}

by solving above equation we will have

B = 2.24 T

5 0
3 years ago
You are playing a game of soccer. Describe as many actionreaction pairs in this situation as you can think of.
marissa [1.9K]
Slide tackle, goalkeeping, avoiding getting tackled, controlling the ball in the air, shooting from a hard angle etc
3 0
3 years ago
Although the class has no water, the students want to know if the cube will float or sink in water. Explain, in detail, the step
Nutka1998 [239]
Okay, so the density of water is 1g/cm3. In order for the cube to float, it has to be less than 1, and it will sink if it is more than 1 g/cm3. Use a triple beam balance to weigh the cube, looking at the metric ruler on the balance. Then, if the cube's density is more than 1, then you know it will float. If the density is less than 1, you know it will sink.
hope this helps, and I didn't know how to use the word "metric ruler"
3 0
3 years ago
in a certain pinhole camera, the screen is 10 cm from the pinhole. When the pinhole is placed 6 cm away from a tree ,a sharp ima
vagabundo [1.1K]

Answer:

Height of the tree is <u>9.6 cm</u>

Explanation:

We know that, Magnification of an image is written as follows.

\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)

Where,

\begin{array}{l}{\mathrm{H}_{0}=\text { height of the object }} \\ {\mathrm{H}_{\mathrm{i}}=\text { height of the image }} \\ {\mathrm{D}_{0}=\text { distance of the object }} \\ {\mathrm{D}_{\mathrm{i}}=\text { distance of the image }}\end{array}

As per given question,

\begin{array}{l}{\mathrm{H}_{1}=\text { height of the image }=\text { height of the image of the tree on screen }=16 \mathrm{cm}} \\ {\mathrm{D}_{0}=\text { distance of the object }=\text { distance of the tree from the pinhole }=6 \mathrm{cm}} \\ {D_{1}=\text { distance of the image }=\text { distance of the image from the pinhole }=10 \mathrm{cm}} \\ {\mathrm{H}_{0}=\text { height of the object }=\text { height of the tree }}\end{array}

Substitute the values in the above formula,

\begin{array}{l}{\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)} \\ {\left(\frac{16}{H_{o}}\right)=\left(\frac{10}{6}\right)} \\ {\mathrm{H}_{\mathrm{o}}=\left(\frac{16 \times 6}{10}\right)} \\ {\mathrm{H}_{\mathrm{o}}=9.6 \mathrm{cm}}\end{array}

Height of the tree is 9.6 cm.

8 0
3 years ago
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