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ArbitrLikvidat [17]
3 years ago
13

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force act

s on the object through the displacement. What is the work done in this case?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

cos 0 = 1.

Fs = 7×8 = 56 J

Explanation:

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A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold
slega [8]

Answer:


F₁ = 4,120.2 N


F₂ = 3,924N


Explanation:



1) Balance of angular momentum around the end where F₁ is applied.


F₂ × 0.5m - F₁ × 0 = mass × g × 1m


⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m


F₂ = 196.2 Nm / 0.5m = 3,924 N


2) Balance of forces


F₁ - F₂ = mg


F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N

4 0
3 years ago
2. If you want 0. 250 a (250 milliamps) to flow around a circuit with a resistance of 400 ohms, what voltage do you need?
grigory [225]

Answer:

0.000625 V

Explanation:

The formula linking current , resistance and voltage is :

V = I/R

Voltage = Current / Resistance

Now we substitute values given in question :

Voltage = 0.250 / 400

Voltage (V) = 0.000625

Our final answer is 0.000625 V

Hope this helped and have a good day

5 0
1 year ago
What happens when two minerals have different arrangements of Atoms
zaharov [31]
The overall arrangements of the atoms produce crystals
5 0
3 years ago
How does being on the moon effect gravitational field strength
SVETLANKA909090 [29]

Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.

However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.

If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.

5 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
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