All categories

2 years ago

Are the small sized gears only force, speed and distance, only speed

1 answer:

3 0

If the gears are of different sizes, they can be used to increase the power of a turning force. The smaller wheel turns more quickly but with less force, while the bigger one turns more slowly with more force. Cars and bicycles use gears to achieve amazing speeds our bodies could never match without help.

You might be interested in

Which of the following is the type of rock

viva [34]

Answer:

d. intrusion

Explanation:

An intrusion is molten rock from the Earth's interior that squeezes into existing rock and cools. Folding Folding occurs when rock layers bend and buckle from Earth's internal forces.

6 0

2 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

2 years ago

Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kÎ©, and

SOVA2 [1]

Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2 + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm

4 0

2 years ago

A 2.1 kg block is dropped from rest from a height of 5.5 m above the top of the spring. When the block is momentarily at rest, t

ella [17]

Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

mass of block, m = 2.1 kg

height above the top of the spring, h = 5.5 m

First, we determine the spring constant based on the principle of conservation of potential energy

¹/₂Kx² = mg(h +x)

¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

U = 2.1 x 9.8 x 5.5 = 113.19 J

Work done in compressing the spring to 15.0 cm:

W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J

This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

K.E = 113.19 J - 42.6 J

K.E = 70.59 J

To determine the speed of the block due to this energy:

KE = ¹/₂mv²

70.59 = ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

v = √67.229

v = 8.2 m/s

8 0

2 years ago

Read 2 more answers

A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax

alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

applying conservation law of momentum

p + .444 i + .748 j = 0

p = - .444 i - .748 j

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0

2 years ago