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Mashcka [7]
2 years ago
11

A frictionless pulley used to lift 8000N of concrete. What is the minimum effort required to raise the block

Physics
1 answer:
rusak2 [61]2 years ago
3 0

Answer: 8000N

Explanation: since it is frictionless that means it has 100% efficiency therefore the mechanical advantage is 1 meaning the load equals to the effort

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Which law is associated with inertia
N76 [4]

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0
3 years ago
A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters
julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0
3 years ago
How do gravitational and electic force compare
velikii [3]

Electric Forces. ... Just like objects that have mass exert gravitational forces on each other, objects that are charged will also exert electric forces on each other. The electric force is directly proportional to the charge of the two objects and inversely proportional to the distance between them squared.

3 0
3 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
3 years ago
What are the strength and direction of an electric field that will balance the weight of a 0.9 g plastic sphere that has been ch
Vladimir79 [104]
(a) The force exerted by the electric field on the plastic sphere is equal to
F=qE
where q=-3.4 nC=-3.4 \cdot 10^{-9} C is the charge of the sphere and E is the strength of the electric field. This force should balance the weight of the sphere:
F=mg =0.9 g
where m is the mass of the sphere and g is the gravitational acceleration.

Since the two forces must be equal, we have:
qE=mg
and so we find the intensity of the electric field
E= \frac{mg}{q}= \frac{0.9 \cdot 9.81 m/s^2}{3.4 \cdot 10^{-9} C} =2.6 \cdot 10^9 N/C

(b) Now let's find the direction of the field. The electric force must balance the weight of the sphere, which is directed downward, so the electric force should be directed upward. Since the charge is negative, the force is opposite to the electric field direction, and so the direction of the electric field is downward.
4 0
3 years ago
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