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Dafna11 [192]
2 years ago
10

Nabil wants to work out the density of a pebble. he has a displacement can and a beaker. What other piece of equipment will he n

eed
Physics
1 answer:
Svetllana [295]2 years ago
3 0

Answer:

1) Measuring cylinder

2) Top Pan Balance

Explanation:

To find the density of the pebble Nabil should first accurately measure the mass of the pebble using the top pan balance

He should then fill the displacement can till the water reaches the outlet pipe at the to of the cylinder

He should then place the pebble into the cylinder carefully ans allow the water that would start coming out to stop

The volume of water in the displaced cylinder is then measured and the density is found by dividing the mass of the pebble by the volume of water in the measuring cylinder

Density \ of \ pebble =  \dfrac{Pebble \ mass}{Volume \ of \ water \ in\ the \ measuring \ cylinder} =  \dfrac{Pebble \ mass}{Water \ volume \ displaced}

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Which statement best compares and contrasts two physical properties of matter?
Stolb23 [73]

Answer:

D

Explanation:

Boiling points and melting points are similar because they both involve the change in a state of a material, but they are different because boiling point involves a change from a liquid to a gas and melting point involves a change from a solid to a liquid.

5 0
2 years ago
A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is
Delicious77 [7]

Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0
2 years ago
Read 2 more answers
An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
Marianna [84]

Answer:

a = 18.28 ft/s²

Explanation:

given,

time of force application, t= 10 s

Work = 10 Btu

mass of the object = 15 lb

acceleration, a =  ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

now,

work done  is equal to change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

 v_f = 182.75\ ft/s

now, acceleration of object

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²

3 0
2 years ago
A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
BlackZzzverrR [31]

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

4 0
2 years ago
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