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3 years ago

A student has a small piece of steel.

Physics

1 answer:

Simora [160]3 years ago

5 0

Answer:

There are different ways to investigate density. In this required practical activity, it is important to:

record the mass accurately

measure and observe the mass and the volume of the different objects

use appropriate apparatus and methods to measure volume and mass and use that to investigate density

Explanation:

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m

just olya [345]

So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c

7 0

2 years ago

Physical form in which a substance can exist

Anika [276]

A solid, liquid, or gas or plasma. Which I think it is. Check though.

4 0

3 years ago

An 800 N man stands on a scale in a motionless elevator. When the elevator begins to move, the scale

DENIUS [597]

Answer:

Net force = 800 – 650

= 150 N

150 = (800 ÷ 9.8) a

a = 1470 ÷ 800

= 1.8375 m/s^2, downwards

3 0

3 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The small object that creates a pivot point is calle

KiRa [710]

Answer:

F = 326.7 N

Explanation:

given data

mass m = 200 kg

distance d = 2 m

length L = 12 m

solution

we know force exerted by the weight of the rock that is

W = m × g ..............1

W = 200 × 9.8

W = 1960 N

and

equilibrium the sum of the moment about that is

∑Mf = F(cos∅) L - W (cos∅) d = 0

here ∅ is very small so cos∅ L = L and cos∅ d = D

so F × L - W × d = 0 .................2

put here value

F × 12 - 1960 × 2 = 0

solve it we get

F = 326.7 N

8 0

3 years ago

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