A 2.5 kg mass starts from rest at point A and moves along the x-axis subject to the potential energy shine in the figure below

elixir [45]

Mass is a measure of how much matter there is within an object, typically given in grams. Mass is not affected by gravity, so a given object would have the same mass on earth as in outer space. Density is the amount of mass in an object per a certain amount of volume

3 0

3 years ago

Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

You are 1.8 m tall and stand 2.8 m from a plane mirror that extends vertically upward from the floor. on the floor 1 m in front

velikii [3]

This problem must be solved using a sketch. I attached an illustration of the problem.
You must trace the ray that reflects from the top off the table to your eyes. This how eyesight works, light rays reflects off the objects into your eyes.
Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
$tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}$
We solve for h:
$\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
2.8h-0.8\cdot 2.8=1.8-h\\
3.8h=1.8+2.24\\
h=\frac{4.04}{3.8}\\
h=1.06m$

6 0

3 years ago

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

20 points and brainliest‼️‼️‼️‼️

Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0

3 years ago