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3 years ago

14

A balloon with a charge of 4.0 x 10 -5C is held a distance of 0.10 m from a second balloon having the same charge. Calculate the

magnitude of the repulsive force.

1 answer:

wolverine [178]3 years ago

3 0

Answer:

1438.4 N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = Electrostatic force on the balloons, k = coulomb's constant, q = charge on the first balloon, q' = charge on the second balloon, r = distance between the two ballons.

From the question,

Given: q = q' = 4.0×10⁻⁵ C, r = 0.1 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( 4.0×10⁻⁵×4.0×10⁻⁵×8.99×10⁹)/0.1²

F = 14.384/0.01

F = 1438.4 N

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A robin in flight has 20.8 J of PE when it is 27.6 m high. What is the mass of the robin? (Unit = kg)

schepotkina [342]

Answer:

<h2> $\boxed{ \bold{ \purple{0.0769 \: kg \: }}}$ </h2>

Explanation:

$\sf{Potential Energy ( P.E ) \: = \: 20.8 \: joule}$

$\sf{distance \: = 27.6 \: metre}$

$\sf{mass = }$ ?

$\sf{acceleration \: due \: to \: gravity = 9.68 \: {metre \: per \: second}^{2} }$

Now, let's find the mass:

$\sf{PE \: = mass \times gravity \: \times \: distance}$

plug the values

⇒ $\sf{20.8 = m \times 9.8 \times 27.6}$

Multiply the numbers

⇒ $\sf{20.8 = 270.48 \: m}$

Swap the sides of the equation

⇒ $\sf{270.48m = 20.8}$

Divide both sides of the equation by 270.48

⇒ $\sf{ \frac{270.48m}{270.48} = \frac{20.8}{270.48} }$

Calculate

⇒ $\sf{0.0769}$ kg

Hope I helped!

Best regards!!

7 0

3 years ago

Read 2 more answers

WILL MARK BRAIN<br> A=154<br> B=145<br> C=26<br> D=206

ANEK [815]

Answer:

154

Explanation:

5 0

3 years ago

50 N of force is exerted on a crate. It accelerates at a rate of 1.2 m/s/s. Calculate its mass

vitfil [10]

Answer:

m = F/a = 50 / 1.2 = 41.6666.... 42 kg

Explanation:

4 0

3 years ago

Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).

Musya8 [376]

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\$

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

$(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}$

<u>u₂ = 3.7 m/s</u>

5 0

2 years ago

3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

3 years ago