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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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Which is true?

Andrew [12]

Answer:

e) A changing magnetic field produces an electric field.

Explanation:

Ok, we start with a magnetic field and let's study how it affects the motion of a single electron. As the magnetic field changes, it will cause an electromotive force, that moves the electron, and because now we have a moving electron, now we will have an electric field. (Such that the direction of the electromotive force opposes the direction in which the magnetic field changes). This also can be deduced if we look at the third Maxwell's equation:

dE/dx = -dB/dt

This says that the spatial change in an electric field depends on how the magnetic field changes as time pass.

Then the correct option is e) A changing magnetic field produces an electric field.

4 0

2 years ago

A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

8 0

2 years ago

State coulomb's law mathematically

Vsevolod [243]

Coulomb's law is express as:

$\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ \text{ where} \\ k\text{ is Coulomb's constant} \\ q_1\text{ and }q_2\text{ are the charges} \\ r\text{ is the distance between charges} \end{gathered}$

7 0

8 months ago

Y’all I have best describes the purpose of a bar graph

Sholpan [36]

so you can see all the different categories at once. both as a whole and on an individual scale.

3 0

3 years ago

What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?

madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0

2 years ago