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Svetradugi [14.3K]
4 years ago
8

A zebra starts from rest and accelerates at 1.9 m/s2 what will be the acceleration after 5 seconds

Physics
1 answer:
Alex73 [517]4 years ago
3 0

Answer:

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

Corrected question;

A zebra starts from rest and accelerates at 1.9 m/s2. How far has the zebra gone after 5 seconds.

Explanation:

From the equation of motion;

d = vt + 0.5at^2 .........1

Where;

d = displacement

v = initial velocity = 0

t = time taken= 5 seconds

a = acceleration = 1.9 m/s^2

Substituting the given values into equation 1;

d = 0(5) + 0.5(1.9×5^2)

d = 0.5(1.9×5^2)

d = 23.75 m

The zebra has gone 23.75m after 5 seconds.

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3 years ago
A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou
qwelly [4]

Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0
3 years ago
You drive a car 1500 ft to the east, then 2500 ft to the north. if the trip took 3.0 minutes, what were the direction and magnit
Lelu [443]

The velocity of the car was 4.93m/s and the direction was northeast.

<h3>Calculation</h3>

It is given,

time of the trip = t = 3 minutes = 180 seconds

1st distance A = 1500ft = 457.2m east

2nd distance B = 2500ft = 762m north

We are asked to find the average velocity.

First, we get the resultant distance magnitude by applying Pythagoras theorem to find the hypotenuse.

A^{2} + B^{2} =C^{2} \\C=\sqrt[]{A^{2} +B^{2} } \\C=\sqrt{457.2^{2} +762^{2} } \\C=  888.63m

Then we have

Velocity = Distance/time

Velocity = 888.63/180

Velocity = 4.93 m/s

Then the direction is northeast.

To know more about Pythagoras theorem, visit:

brainly.com/question/343682

#SPJ4

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2 years ago
A space probe is built with a mass of 1700 pound-mass [lbm] before launch on Earth. The probe is powered by four ion thrusters,
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mass of the space probe is given as

m = 1700 lbm = 771.12 kg

each thrust will apply a net force

F = 225 milli N = 0.225 N

now we will have

v = at

v = \frac{F}{m}*t

given that we have final speed v = 420 miles/minute

we need to convert it into m/s

v = 420\frac{miles}{min}*\frac{1609 m}{1 miles}*\frac{1 min}{60 s}

v = 11263 m/s

now from above equation

11263 = \frac{0.225}{771.12}* t

t = 3.86 * 10^7 s

t = 10722.3 hours = 446.8 days = 63.8 weeks

<em>so it will require 63.8 weeks to reach the given speed</em>

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