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OlgaM077 [116]
4 years ago
7

What tool is used to determine the mass of an object?

Physics
1 answer:
melisa1 [442]4 years ago
3 0
A Beam balance (or Beam scale) is a device to measure weight or mass. These are also known as mass scales, weight scales, mass balances, weight balances, or simply scales, balances, or balance scales. You can also use a graduated cylinder and determine the water displaced and figure out the mass of an object this way!

- if this answer is what you were looking for, don’t forget to heart and maybe if you really liked it brainliest :)
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True or false? The plates keep moving and are still moving today.
wariber [46]

Answer:

true

Explanation:

takes time for most plates to move but throughout time they keep moving just at a slow pace

6 0
3 years ago
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
4 years ago
5. One electrical appliance operates with a voltage of 120 V, while another operates with 240 V. Based on this information alone
serious [3.7K]

Answer:

Yes it is true saying that 2nd applience use more power than first one

4 0
3 years ago
People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th
Whitepunk [10]

Answer:

Current needed = 704A

Explanation:

Using the fomula; torque(τ) = (I)(A)(B)Sinθ

Where B = uniform magnetic field

I = current and A = Area

Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m

Area(A) = πr^(2) = πr^(2)

= π(0.095)^(2) = 0.0284 m^(2)

Now, B(earth)= 5x10^-5 T

While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.

Now, if we arrange the formula to solve for charge (I):

I = (τ)/(A)(B)

I = (1.0x10^-3) / (0.0284)(5x10^-5)

I = 704A

6 0
3 years ago
After the layers of sediment are deposited in one place, they are
myrzilka [38]

your question doesnt make sense

5 0
4 years ago
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