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Rudiy27
3 years ago
6

When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire? A. Increases the force of gravi

ty B. Increases the force of friction C. It decreases the force of friction D. It decreases the force of gravity
Physics
2 answers:
larisa86 [58]3 years ago
8 0
Question: <span>When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire?

C. It decreases the force of friction.

First mud is not a solid. It is more of a slow liquid like corn syrup. And when you are caught in mud or trying to ride past the mud you loose friction with the ground.
You could also resemble friction with traction. In a car. If there is water & oil mixed on the road and you go past it at a high mph range. You will loose traction and start skipping against the mixed substance. This reduces your traction (friction) with the ground.

Final Answer:

C. It decreases the force of friction
</span>
Flura [38]3 years ago
7 0
The answer to this is C.
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A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

8 0
3 years ago
Read 2 more answers
What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma
Sindrei [870]

Answer:

E = 1.50 × 10^{8} V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

\mu _b = \frac{B}{2\mu _o}   ...............1

and

energy density due to electric filed is

\mu _e = \frac{\epsilon _o E^2}{2}     ...............2

and here \mu _b = \mu _ e

so that

E = \frac{B}{\sqrt{\mu _o \times \epsilon _o}}      ...................3

put here value and we get

E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}  

E = 3 × 10^{8}  × 0.50

E = 1.50 × 10^{8} V/m

6 0
3 years ago
Please. Physics is so difficult.
Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

0.984 - 0.994 = 0.010 m

4 0
2 years ago
Read 2 more answers
Why are many mitochondria needed in cells that move, like muscle cells?
Deffense [45]
Muscle cells have to use a lot of energy and the Mitochondria is known as the "power house" of the cell... I would say the number of mitochondria needed is increased because more power and energy is being used in that cell than a normal one (idk if this helps or if that's the answer you were looking for)
3 0
3 years ago
A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f
Rufina [12.5K]

Answer:

Kf = 470 mJ

Explanation:

  • According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
  • Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
  • Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

       W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)

  • As we have already said, (1) is equal to the final kinetic energy of the puck:
  • ⇒ Kf = 470 mJ  (2)
8 0
3 years ago
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