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A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

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Answer 1

Answer:

Question:

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

Statement:

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

Solution:

Initial velocity (u) = 70 m/s
Acceleration (a) = -14 m/s^2
Time (t) = 3 s
Let the velocity of the car after 3 s be v m/s
By using the formula,

v = u + at, we have

$v = 70 + ( - 14)(3) \n = > v = 70 - 42 \n = > v = 28$

So, the velocity of the car after 3 s is 28 m/s.

Answer:

The car's speed after 3 s is 28 m/s.

Hope it helps

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Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?
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Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.

Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer:

-9Q

Explanation:

Electric field at origin is:

$E=(2keQ)/(a^2)$

Electric field due to first charge at origin would be:

$E_1=(keQ)/(a^2)$

Electric field due to second charge would be:

$E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)$

If the second charge is Q', then $E_2$ should be:

$E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)$

compare the above two values to find the possible values of Q':

$(|Q'|)/(9)=Q\n |Q'|=9Q$

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.

Learn more about Electric field here:

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An electromagnetic wave is traveling the +y direction. The maximum magnitude of the electric field associated with the wave is Em, and the maximum magnitude of the magnetic field associated with the wave is Bm. At one instant, the electric field has a magnitude of 0.25 Em and points in the +x direction. At this same instant, the wave's magnetic field has a magnitude which is ... and it is in the ... direction.

Answers

Answer:

0.83x10^-9 T

Direction is towards +z axis.

Explanation:

E = cB

E = magnitude of electrical 0.25 Em

c = speed of light in a vacuum 3x10^8 m/s

Therefore,

B = E/c = 0.25 ÷ 3x10^8

B = 0.83x10^-9 T

Magnetic fueld of a EM wave acts perpendicularly to its electric field, therefore it's direction is towards the +Z axis

A train travels due south at 20 m/s. It reverses its direction and travels due north at 20 m/s. What is the change in velocity of the train? 50 m/s, due south e50 m/s, due north 120 m/s, due south zero ms 40 m/s, due north

Answers

Answer:

40 m/s due north

Explanation:

Consider that the south direction a negative Y axis and north direction as + Y axis

v1 = 20 m/s South = 20 (-j) m/s

v2 = 20 m/s North = 20 j m/s

Change in velocity = v2 - v1 = 20 j - 20 (-j) = 40 j m/s

So, change in velocity is 40 m/s due north.

A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °

Answers

Answer: $\theta =45.73^(\circ)$

Explanation:

Given

Length of string =2.15 m

mass of ball =5.49 kg

speed of ball=4.65 m/s

Here

Tension provides centripetal acceleration

$T\cos\theta =mg$ -----1

$T\sin \theta =(mv^2)/(r)$ ------2

Divide 2 & 1

$tan\theta =(v^2)/(rg)$

$tan\theta =(4.65^2)/(2.15* 9.8)$

$tan\theta =1.026$

$\theta =45.73^(\circ)$

If the velocity of a pitched ball has a magnitude of 41.0 m/s and the batted ball's velocity is 50.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answers

The change in momentum is (91 m/s) multiplied by the mass of the ball (which you neglected to mention).
That's exactly the impulse delivered by the bat.

The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m , what was its initial speed v ?.

Answers

The given situation is illustrated below. A particle is released and given a quick push. As a result, it acquires a speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, the initial speed of the particle is $√((2qV/m).)$

To solve the problem, we need to apply the law of conservation of energy, which states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
Initial potential energy = Final kinetic energy
The initial potential energy of the particle is given by
U = qV
where V is the potential difference between the corner and the center of the square.
At the center of the square, the potential energy is zero.
The final kinetic energy of the particle is given by
K = (1/2) mv^2
where m is the mass of the particle and v is its final velocity.
Since the particle is momentarily at rest at the center of the square, its final kinetic energy is zero.
Therefore, we have
qV = (1/2) mv^2
Solving for v, we get
v = $√((2qV/m).)$

for such more question on speed

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