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r-ruslan [8.4K]
3 years ago
7

An object is electrically charged if the amounts of positive and negative charge it contains are not _______.

Physics
2 answers:
Naddik [55]3 years ago
4 0
Equal bc it’s neutral
zlopas [31]3 years ago
3 0

Answer:oops

Explanation:

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Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from
svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2   force of Sun on Mars

F(Merc) = G m M / R^2    force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0
2 years ago
Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc
Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

m_1 and m_2

so we will have

Force applied on box 1 then acceleration

a = \frac{F}{m_1 + m_2}

1.50 = \frac{12.2}{m_1 + m_2}

m_1 + m_2 = 8.13

Now we know that contact force between them in above case is given as

F_n = m_2a

8.90 = m_2(1.5)

m_2 = 5.93 kg

now we have

m_1 = 2.20 kg

8 0
2 years ago
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Lemur [1.5K]

Explanation:

Work cannot be increased by using a machine of some kind.

8 0
2 years ago
If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object
12345 [234]

Answer:

ΔK = 24 joules.

Explanation:

ΔK = Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

W = F * Δs

Δs can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δs = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

W = < 3, 4 > * < 4, 3 > = (3*4)+(4*3) = 24 J

3 0
1 year ago
Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r
IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance

V is the potential difference

Calling C_1 the capacitance of capacitor 1 and V_1 its potential difference, the energy stored in capacitor 1 is

U=\frac{1}{2}C_1 V_1^2

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: C_2 = \frac{C_1}{2}

- The voltage is twice the voltage of capacitor 1: V_2 = 2 V_1

so the energy stored in capacitor 2 is

U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2

So the ratio between the two energies is

\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}

4 0
2 years ago
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