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3 years ago

A uniform brick of length 21 m is placed over

Physics

1 answer:

Paha777 [63]3 years ago

6 0

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

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If a force of 30N is applied to an object of mass 8kg which is kept on a surface, it achieves an acceleration of 3m/s. What is t

Eduardwww [97]

By Newton's second law, the net force on the object acting parallel to the surface is

∑ F = F[applied] - F[friction] = (8k g) (3 m/s²)

If F[applied] = 30 N, then

30 N - F[friction] = 24 N ⇒ F[friction] = 6 N

so the answer is B.

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2 years ago

The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a sp

AlekseyPX

Answer:

$t = 0.93 s$

Part b)

$d = 3.98 m$

Part c)

$v_x = 4.28 m/s$

$v_y = -11.49 m/s$

Explanation:

The two components of the velocity of the ball is given as

$v_x = 4.9 cos29$

$v_x = 4.28 m/s$

$v_y = 4.9 sin29$

$v_y = 2.37 m/s$

Part a)

now we know that the displacement in y direction is given as

$\Delta y = 6.4 m$

so we have

$\Delta y = v_y t + \frac{1}{2}gt^2$

$6.4 = 2.37 t + 4.9 t^2$

$t = 0.93 s$

Part b)

Distance of the ball in x direction of the motion is given as

$d = v_x t$

$d = 4.28 \times 0.93$

$d = 3.98 m$

Part c)

In x direction the velocity will remain the same always

$v_x = 4.28 m/s$

while in Y direction we can use kinematics

$v_y = v_{oy} + at$

$v_y = -2.37 - 9.81(0.93)$

$v_y = -11.49 m/s$

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2 years ago

Two charged particles are placed 2.0 meters apart. the first charge is +2.0 e-6 c and the second charge is +4.0 e-6

Yakvenalex [24]

The Coulomb force between two charges is defined as: Fc=(k*Q₁*Q₂)/r² where k=9*10^9 N m² C⁻², Q₁ and Q₂ are charges and r is the distance between those charges.
In our case:

Q₁=2*10^-6 C
Q₂=4*10^-6 C
r=2 m

Now we simply plug in the numbers into the equation:

Fc={(9*10^9)*(2*10^-6)*(4*10^-6)}/2^2=0.018 N

The Coulombs force between the two positive charges is Fc=0.018 N and it is a repulsive force because both charges are positive.

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3 years ago

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Lesechka [4]

Answer:

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2 years ago

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