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stellarik [79]
2 years ago
10

Gizmo Warm-up: Lifting a piano A pulley is a simple machine that is used to lift heavy objects. A pulley is a wheel with a groov

e for a rope. Pulling down on one end of the rope causes the other end to pull up. In the Object to lift menu choose Piano. What is the weight of the piano
Physics
1 answer:
Nat2105 [25]2 years ago
5 0

Answer:

bruuuuuuuuh

Explanation:

id                  

You might be interested in
In which of the following is positive work done by a person on a suitcase
MrRa [10]

pendajo sMh so stupid

4 0
3 years ago
You can keep a 1kg apple from falling to the ground by placing it on a table. What reaction force is resisting the force of the
Gnom [1K]

The equilibrium condition allows finding the correct answer for the force that is resisting the weight of the apple is:

        3. Normal force

Newton's second law gives the relationship between <em>force, mass</em> and acceleration of bodies, in the special case that the acceleration is is called the equilibrium condition.

            ∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a scheme of the forces.

Let's write the equilibrium condition for the apple

          N - W = 0

          N = W

.

We can see that the only forces acting on the apple are its weights and reaction from the table called Normal.

Let's analyze the different answers:

1. False. The apple is not moving therefore the resistance is zero

2. False. The apple is not moving  so friction it with the table is  

3. True. The free-body diagram shows that the normal and the weight are equal

4. False. There is nothing to pull the apple so there is no tension

5. False. Gravity is the weight of the apple that is applied to the table, not from the table to the apple.

In conclusion using the equilibrium condition we can find the correct result for the force that is resisting the weight of the apple is:

        3. Normal force

Learn more here: brainly.com/question/2872207

8 0
3 years ago
Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c
Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0
3 years ago
What is demonstrated by water moving up a straw?
Alja [10]
<span>The answer is C: water is drawn up a straw by cohesion and adhesion. Water molecules stick to one another and the walls of the straw, just like in a capillary.
Cohesion is the attractive force between like materials (between water
molecules).
Adhesion is the attractive force between twounlike materials (such as between
water and a solid container).
Capillary action is the tendency of a liquid to rise innarrow tubes or small openings as a result of adhesion and cohesion.
The liquid water molecules bind to the straw—a process known as adhesion.  In the narrow space of the straw, the interaction of cohesion and adhesion causes theliquid to be drawn upward in the straw.</span>
8 0
3 years ago
Read 2 more answers
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

6 0
3 years ago
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