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BabaBlast [244]

3 years ago

11

A heavy bank-vault door is opened by the application of a force of 300 N directed perpendicular to the plane of the door at a di

stance of 0.80 m from the hinges. What is the torque?
300 Nm

240 Nm

375 Nm

120 Nm

1 answer:

kondaur [170]3 years ago

5 0

The answer is 240 Nm.

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There would be 6 electrons placed on the third energy level.

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Explain why squeezing a plastic mustard bottle forces mustard out to top.

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What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

$F=k\dfrac{q_1q_2}{d^2}$

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

$F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F$

So, the new force becomes 4 times the initial force.

4 0

3 years ago

Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction

andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

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3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

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3 years ago