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drek231 [11]
3 years ago
7

You rub a rod of glass with a cotton cloth, then dip the cloth into a Faraday pail like we will be using in lab. Will a charge s

ensor connected to the pail give a negative reading, or a positive reading or will it read essentially zero?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:

negative charge

Explanation:

As the glass rod is rubbed with the cotton cloth, then the glass rod becomes positively charged and the cotton cloth becomes negatively charged.

There is a tendency to lose the electrons by the glass rod and to grab the electrons by the cotton cloth.

So, the sensor shows negative charge.

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A food scientist often develops new __________________________.
Cerrena [4.2K]

Answer:

spices. i think.

Explanation:

B spices

6 0
3 years ago
Read 2 more answers
Question number 6 :<br><br> Is my answer right?? Plzz help
Dennis_Churaev [7]
No.  You have the units wrong.

Example with numbers: 

When you multiply  (1/9) by 2, you get 2/9 .

With units:

When you multiply  (m/s) by (kg), you get (kg-m/s).
3 0
3 years ago
A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067
Vesna [10]

Answer:

The mass of the bullet is 15 g.

(b) is correct option.

Explanation:

Given that,

Momentum = 2.8 kg m/s

Speed = 187 m/s

We need to calculate the mass of the bullet

Using formula of momentum

P = mv

m = \dfrac{P}{v}

Where, P = momentum

v = speed

Put the value into the formula

m = \dfrac{2.8}{187}

m = 0.015\ kg

m = 15\ g

Hence, The mass of the bullet is 15 g.

4 0
3 years ago
A block with a mass of 0.600 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilib
tankabanditka [31]

Answer:

Explanation:

The amplitude of the oscillation under SHM will be .5 m and the equation of

SHM can be written as follows

x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.

x = .5 cosωt

given , when t = .2 s , x = .35 m

.35 = .5 cos ωt

ωt = .79

ω = .79 / .20

= 3.95 rad /s

period of oscillation

T = 2π / ω

= 2 x 3.14 / 3.95

= 1.6 s

b )

ω = \sqrt{\frac{k}{m} }

ω² = k / m

k = ω² x m

= 3.95² x .6

= 9.36 N/s

c )

v = ω\sqrt{(a^2-x^2)}

At t = .2 , x = .35

v = 3.95 \sqrt{.5^2-.35^2}

= 3.95 x .357

= 1.41 m/ s

d )

Acceleration at x

a = ω² x

= 3.95 x .35

= 1.3825 m s⁻²

7 0
3 years ago
What pattern is seen on the moon from earth?
igomit [66]

If you mean pattern designs it’s rocky bumpy and you can see the circle patterns

If you mean phases the here’s a picture

7 0
3 years ago
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