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weqwewe [10]
3 years ago
9

A 357 kg box is pulled 8.00 m up a 30° frictionless, inclined plane by an external force of 4925 N that acts parallel to the pla

ne. Calculate the work done by the external force. work done by the external force: J Calculate the work done by gravity. work done by gravity: J Calculate the work done by the normal force.
Physics
1 answer:
Sidana [21]3 years ago
4 0
I'm so tremble right now
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A positron has a mass of 9.11 x 10^-31 kg, and charge qp = +e = +1.60 x 10^-19 C. It is moving towards an α particle (qα = +2e,
OlgaM077 [116]

Answer:

Part a)

v = 2.25 \times 10^6 m/s

Part b)

r = 0.72 \times 10^{-10} m

Explanation:

As we know that alpha particle remain at rest always so the energy of system of positron and alpha particle will remain constant

So we will have

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

here we know that

q_1 = 1.6 \times 10^{-19} C

q_2 = 2(1.6 \times 10^{-19}) C

also we have

r_1 = 2.00 \times 10^{-10} m

r_2 = 1.00 \times 10^{-10} m

v_1 = 3.00 \times 10^6 m/s

m = 9.11 \times 10^{-31} kg

now from above equation we have

2.304 \times 10^{-18} + 4.0995\times 10^{-18} = 4.608 \times 10^{-18} + \frac{1}{2}(9.11 \times 10^{-31})v^2

v = 2.25 \times 10^6 m/s

Part b)

Now when it come to rest then again by energy conservation we can say

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now here final speed will be zero

2.304 \times 10^{-18} + 4.0995\times 10^{-18} = \frac{(9/times 10^9)(1.6\times 10^{-19})(2\times 1.6 \times 10^{-19})}{r_2}

r = 0.72 \times 10^{-10} m

7 0
3 years ago
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on
nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

8 0
3 years ago
Which of the following statements is falsifiable?
dimulka [17.4K]

We have that the <em>statements</em> found falsifiable is

Henry Ford invented the first automobile.

Option A

<h3>Henry Ford</h3><h3 />

Generally in as much Ford motor is a great vehicle company

He did not create the first automobile

Credit is give to Karl Benz for The first auto-mobile

Therefore

The statements found falsifiable is

Henry Ford invented the first automobile.

Option A

For more information on Benz visit

brainly.com/question/19424882

7 0
3 years ago
A football (450 g) falls vertically from rest onto the ground from a height of 5.0 m. If the force that
mario62 [17]
I don’t understand sorry
5 0
3 years ago
What is the speed of a bird of mass 8kg which has kinetic energy of 8836J?
Kay [80]

Answer:

For the bird moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: Ek=12mu2.

4 0
3 years ago
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