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3 years ago

A beaker contain 200mL of water What is its volume in cm3 and m3

Physics

2 answers:

Bogdan [553]3 years ago

7 0

The volumes are 200cm3 and 0.0002m3

Nutka1998 [239]3 years ago

4 0

A mL is equal to a cm^3, so it is 200 cm^3. To help you transfer cm^3 to m^3, use KHDBDCM. You move the decimal point twice to the left which means that it is 2 m^3.
TLDR: 200 cm^3 and 2 m^3.

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Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.

AfilCa [17]

Answer:

P.E. = -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

P.E. = -0.449 J

8 0

3 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

djyliett [7]

Answer:

a) m = 69.0 kg

b) release some gas in the opposite direction to the astronaut's movement

Explanation:

a) Let's use Newton's second law

F = m a

m = F / a

m = 60.0 / 0.870

m = 69.0 kg

b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved

p₀ = p_f

m v = M v '

v ’= $\frac{m}{M} \ v$

so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.

One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.

3 0

3 years ago

A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

snow_lady [41]

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

7 0

3 years ago

What are 3 types of contact forces

Svetach [21]

There are different types of contact forces like normal Force, spring force, applied force and tension force.

5 0

3 years ago

Read 2 more answers

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago