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3 years ago

Why do you keep moving forward when you slam on the brakes of your bike?

2 answers:

7 0

If you're going really really slow and you hit the brakes, you're going to stop almost instantly. When you're going faster, you've got your energy which is your body size so it takes energy to stop the bike. When you slam the brake you're trying to apply that energy to the wheels. Even if you lock them up so the wheels aren't spinning, you skid because friction is rubbing against the ground and that will be what eventually stops you.

If you have a wider tire, you're going to have more surface area hitting the ground and that will cause more friction so you will stop faster. If you have skinnier tires you will skid more so there are a bunch of different factors.

g100num [7]3 years ago

6 0

Because the act of braking is an example of negative acceleration.
Example: if the rate of braking was say 2 meters per second^2, and the starting velocity was 10 m/s, it would take 5 seconds to come to a stop(during those 5 seconds you would still be moving).

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In the 19th century James Joule, an English scientist, was the first to recognize the mechanical equivalent of heat as a specifi

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Answer:

D) wood rubbed against a rough surface feels hot

Explanation:

The heat is transferred from one form of energy (friction of the wood being rubbed against the surface) to another (heat energy).

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3 years ago

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Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe

lutik1710 [3]

Answer:

From the formula of force:

$F = \frac{kAB}{ {r}^{2} } \\$

since AB and k are constants:

$F \: \alpha \: \frac{1}{ {r}^{2} } \\ \\ F = \frac{x}{ {r}^{2} }$

x is a constant of proportionality

• when force is 4N, separation distance is 1

$4 = \frac{x}{1} \\ x = 4$

therefore, equation becomes

$F = \frac{4}{ {r}^{2} } \\$

when r is doubled, r becomes 2. find F:

$F = \frac{4}{ {2}^{2} } \\ \\ F = \frac{4}{4} \\ \\ { \underline{force \: is \: 1N}}$

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3 years ago

A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor

xxMikexx [17]

The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v is the initial velocity of the jumper on the floor
h is the maximum height reached by the jumper
g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
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1 year ago

Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.

natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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2 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$