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3 years ago

how much force is needed to cause a 15 kilogram bike to accelerate at a rate of 10 meters per second?

Physics

2 answers:

egoroff_w [7]3 years ago

8 0

F = m*a, mass times acceleration.

F = 15*10 = 150 N

Gwar [14]3 years ago

6 0

Answer:

150 Newtons

Explanation:

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Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

A student releases a marble from the top of a ramp. The marble increases

Fed [463]

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at

Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)

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3 years ago

When approaching an intersection, bridge, or railroad crossing, you should never drive (pass on the left half of the roadway whe

e-lub [12.9K]

You should not go into the left side of the roadway when within 100 feet of the crossing. Moreover, you should also turn on your turn signal when within 100 of a turn. These precautions prevent accidents as it makes clear to other drivers what your intentions are and drivers making turns are not endangered.

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3 years ago

What happens if :<br> . The test charge is not tiny.

docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

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2 years ago

The sun is 150,000,000 km from the earth.

aniked [119]

Answer:

<h3>What is the angular speed of the earth around the sun? </h3>

It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of

$7.2921159 × 10 {}^{ - 5} \: \: \frac{radians}{second}$

<h3>What is the tangential speed of the earth? </h3>

The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.

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3 years ago