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serg [7]
3 years ago
6

Select the correct answer.

Physics
1 answer:
Bess [88]3 years ago
3 0

Answer:

C

Explanation:

V-V₀=at

a=(V-V₀)/t

a=(10-20)/5=-2 m/s²

Acceleration of the car is -2 m/s²

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What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As
Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

8 0
3 years ago
Find the potential energy of a 2 kg ball 15 m in the air.
mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

8 0
3 years ago
The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0
2 years ago
A car of mass 998 kilograms moving in the positive y–axis at a speed of
aksik [14]
The total momentum should come out to be  <span>2.0 x 10^4 kilogram meters/second </span>
8 0
3 years ago
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