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2 years ago

How much support force does a table exert on a book that weighs 15 N when the book is placed on the table?

Physics

1 answer:

mestny [16]2 years ago

8 0

Answer:

15 N

Explanation:

According to Newton's third law of motion, to every action, there is an equal and opposite reaction. This reaction is equal in magnitude to the force acting but in an opposite direction.

Now, if the book weighs 15 N, an opposite equal force will be: N = -15 N

But the magnitude of this will be the absolute value which is 15N.

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Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that th

lisabon 2012 [21]

Answer:

Using equation

qs=F

q=VQ/I V=1200

q=998.45 N

qs=F

s=F/q

s=700/998.45

spacing=s=0.701 m

7 0

3 years ago

Read 2 more answers

Why can't you hear a sound in space ?

Anastaziya [24]

Answer:

As the vibrations of the particles reach your ear, your ear drum receives the vibrations which the brain then interprets as sound. In the vacuum of space, there are no (or very, very few) particles to vibrate, so sound cannot travel through this medium.

Explanation:

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2 years ago

According to Coulomb's law, what happens to the attraction of two oppositely charged objects as their distance of separation inc

NARA [144]

Answer:

Option B. Decreases

Explanation:

Coulomb's law states that:

F = Kq₁q₂ / r²

Where:

F => is the force of attraction between two charges

K => is the electrical constant.

q₁ and q₂ => are the two charges

r => is the distance apart.

From the formula:

F = Kq₁q₂ / r²

The force of attraction (F) is inversely proportional to the square of their separating distance (r).

This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.

Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.

Option B gives the right answer to the question.

7 0

3 years ago

An apple drops from the tree and falls freely. The apple is originally at rest a height H0 above the ground. The ground is cover

zzz [600]

Answer:

Explanation:

The distance travelled in the free fall is H - h

Since the apple originally started from rest we can use v^2 = u^2 + 2 x g x s where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0

So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]

Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.

So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]

so since v^2 = u^2 + 2 x a x s then

O^2 = 2 x g x (H-h) + 2 x a x h

{O^2 - 2 x g x (H - h)}/(2 x h) = a

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3 years ago

A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o

Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0

3 years ago