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3 years ago

Why does friction increase as speed increases??

1 answer:

7 0

Friction is directly related to air particles.

When we say that friction is high, it means that you're colliding with lots of air particles, and hence you can't speed up as easily.
Thus, the more air particles you encounter, the higher the friction.

The faster you go, the more particles you will encounter in a given time; hence at higher speeds, the friction is higher.

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A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t

WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

$v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s$

So, the required time is 2.2 seconds.

3 0

2 years ago

A green plant absorbs light a frog eats flies these are both examples of how organisms

Llana [10]

I think I can answer your question since I've worked on this before. Your answer should be obtain energy. If your answer choices were; obtain energy escape predators produce offspring excrete waste

7 0

2 years ago

A bowling ball with a negative initial velocity slows down as it rolls down the lane toward the pins. Is the bowling ballâs acce

stiv31 [10]

Answer:

positive

Explanation:

The ball is rolling down with a negative velocity, but the velocity is slowing down. therefore the velocity must increase in order for the ball to slow down.

For example let the ball's initial velocity be -15 m/s. and it is slowing down to let's say -13 m/s. Well this means that it's velocity has increase by 2 m/s. So, its acceleration is positive.

5 0

2 years ago

Ballistic data obtained on a firing range show that aerodynamic drag reduces the speed of a .44 magnum revolver bullet from 250

m_a_m_a [10]

Answer:

0.363999909622

Explanation:

F = Force

m = Mass = 15.6 g

C = Drag coefficient

ρ = Density of air = 1.21 kg/m³

A = Surface area = $\dfrac{\pi}{4}d^2$

v = Terminal velocity = $v=210\ m/s$

s = Displacement = 150 m

$a=\dfrac{v^2-u^2}{2s}$

Force is given by

F = ma

$F=\dfrac{1}{2}\rho CAv^2\\\Rightarrow ma=\dfrac{1}{2}\rho CAv^2\\\Rightarrow m\dfrac{v^2-u^2}{2s}=\dfrac{1}{2}\rho CAv^2\\\Rightarrow C=2\times m\dfrac{v^2-u^2}{2s}\times\dfrac{1}{\rho Av^2}\\\Rightarrow C=2\times15.6\times 10^{-3}\dfrac{210^2-250^2}{2\times 150}\times\dfrac{1}{1.21\times\dfrac{\pi}{4}\times (11.2\times 10^{-3})^2(210)^2}\\\Rightarrow C=-0.363999909622$

The drag coefficient is 0.363999909622 (ignoring negative sign)

4 0

3 years ago

A force of 3,200 kg x m/s^2 (Newton’s) acts on a truck giving it an acceleration of 2 m/s^2. What is the mass of the truck ?

Ivan

Answer:

According to newton's second law of motionF=ma Data:-F=3200kgm/sec² or N ,a=2m/sec² ,m=? solution :-F=ma here we have to find m so m=F/a ,m=3200/2=1600kg

3 0

3 years ago