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Nitella [24]
2 years ago
6

Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of

mass m1, in meters per square second.

Physics
1 answer:
eduard2 years ago
7 0

Answer:

a1 = 3.68m/s²

Explanation:

Given values:

Mass of the block placed on the table, m1 = 12.25 kg

Mass of the block hanging vertically, m2 = 7.5 kg

Acceleration due to gravity, g = 9.8 m/s2

Tension in the string is T

Let the acceleration of mass 1 and mass 2 be a1 and a2

a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,

a = m2*g/(m1 + m2)

a = 7.5 x 9.8 / (12.5 + 7.5)

a = 3.68 m/s²

a1 = a2 = 3.68m/s²

a1 is directed to the right and a2 is directed downwards

Below is a diamonds to show the geometrical arrangements of both masses

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An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the
Nataly [62]

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

F=ke

m is the mass of the block = ?

a is the acceleration = 0.8m/s²

e is the extension of the spring =  0.88m

k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

m=49.5kg

The mass of the object is 49.5kg which is approximately 50kg

6 0
2 years ago
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The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the
Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

m = IS

Where,

I = Current

S = Area

m = IA \rightarrow m= I( \pi r^2)

Replacing with our values we have that,

8*10^{22} = I \pi(\frac{3000*10^3}{2})^2

Re-arrange to find I,

I = 1.1317*10^{10} A

Part B) To find the Current density we need to find the cross sectional area of the Wire:

A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2

Finally the current density is simply J

J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

A = \pi (10-3)^2 \\A = 49\pi

Therefore the current density would be

J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2

Comparing the two values we can see that the 2mm wire has a higher current density.

4 0
1 year ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

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k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

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