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4 years ago

How does a bed exert force on you

1 answer:

8 0

For every force, there is an equal and opposite reactionary force.

So when your weight acts on the bed, in order for the bed to not collapse, it must be able to exert an equal and opposite force back on you.

You might be interested in

How much work does a 60kg person do in walking up 5m of stairs

Nataliya [291]

Answer:

If I’m correct 300 joules

Explanation:

8 0

3 years ago

How can we realize that light travel in straight line ?

Norma-Jean [14]

Answer:

It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.

Explanation:

The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.

One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.

3 0

3 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

Svetlanka [38]

Answer:

An electrical current

Explanation:

An electrical current

4 0

3 years ago

A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the

siniylev [52]

Answer:

$K=58.8N/m$

Explanation:

From the question we are told that:

Mass $M=0.442$

Drop distance $d=0.150$

Generally the equation for Spring Constant is mathematically given by

$K=\frac{2mg}{x}$

$K=\frac{2*0.442*9.8}{1.150}$

$K=58.8N/m$

6 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago