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3 years ago

The source of the centripetal force that arises when a runner rounds a curve on a track is _____.

2 answers:

5 0

while taking a turn on a track, the track provides the frictional force acting towards the center of the curve. we know that to move in a curve, centripetal force is required. here since the frictional force acts towards the center, it provides the necessary centripetal force to move in circle. hence the correct choice is

friction

mel-nik [20]3 years ago

3 0

It's actually Friction.

I just did the test and got it right.

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Complete the sentences below using the words below, you may use each word more than once

satela [25.4K]

Decreases, stays the same, increases. The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume). The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules. The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.

7 0

3 years ago

Need help solving this

Nostrana [21]

Answer:

cant see picture

Explanation:

5 0

3 years ago

When Earth and the Moon are separated by a

Wewaii [24]

Using the Universal Gratitation Law, we have:

$F= \frac{MmG}{d^2} \\ MmG=2*10^{20}*(3.84*10^8)^2 \\ MmG=29.4912*10^36$

Again applying the formula in the new situation, comes:

$F= \frac{MmG}{d^2} \\ F= \frac{29.4912*10^36}{(1.92*10^8)^2} \\ \boxed {F=8*10^{20}}$

Number 4

If you notice any mistake in my english, please let me know, because i am not native.

6 0

3 years ago

Read 2 more answers

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

An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric

8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0

3 years ago