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2 years ago

1mm is equal to .1 what

Physics

1 answer:

bulgar [2K]2 years ago

8 0

Answer:

When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.

1 foot (ft) = 0.3048 meters (m)

BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:

1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m

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A child is sitting on the seat of a swing with ropes 5 m long. Her father pulls the swing back until the ropes make a 30o angle

morpeh [17]

Answer:

v = 3.7 m/s

Explanation:

As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:

m. g. h = 1/2 m v²

The only unknown (let alone the speed) in the equation , is the height from which the swing is released.

At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.

As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:

h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m

Replacing in the energy conservation equation, and solving for v, we get:

v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s

7 0

3 years ago

Determine the electrical and gravitational forces that two protons in the nucleus of a helium atom exert on each other when sepa

jonny [76]

Answer:

Fe = 25.67 N

Fg = 2.0734 x 10^-35 N

Explanation:

r = 3 x 10^-15 m

G = 6.67 x 10^-11 Nm^2/kg^2

Mp = 1.6726231 x 10^-27 kg

Qp = 1.6021 x 10^-19 C

K = 9 x 10^9 Nm^2/C^2

The formula for the electrical force between the two protons is given by

$F = K \frac{Q_{p}\times Q_{p}}{r^{2}}$

$F = \frac{9\times 10^{9}\times 1.6021\times 10^{-19}\times 1.6021\times 10^{-19}}{3\times 10^{-15}\times 3\times 10^{-15}}$

Fe = 25.67 N

The formula for the gravitational force between the two protons is given by

$F = G \frac{M_{p}\times M_{p}}{r^{2}}$

$F = \frac{6.67\times 10^{-11}\times 1.6726231\times 10^{-27}\times 1.6726231\times 10^{-27}}{3\times 10^{-15}\times 3\times 10^{-15}}$

Fg = 2.0734 x 10^-35 N

4 0

3 years ago

Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuu

zlopas [31]

Answer:

Explanation:

we know that

s=vt here v is the speed and s is distance covered by the signals

given data

v=3*10^8

t=10 min we have to convert it into seconds

1 minute=60 seconds

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

8 0

3 years ago

Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c

mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

$V = \frac{k*q}{r}$

where k = $\frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2$

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

$V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)$

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

$d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}$

Replacing by the values in (1) we have:

$V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q) = 0 V$

which is equal to the option d).

6 0

3 years ago

You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located

Natalija [7]

To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

$v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})$