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xxMikexx [17]
3 years ago
14

a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired

be the distance trouble​
Physics
1 answer:
blondinia [14]3 years ago
6 0
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
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this event is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the oc
MissTica

Answer: Tsunami.

Explanation :

Tsunami is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the ocean floor.

Tsunami is caused due to the displacement of a large volume of water like in an ocean. It consists of a series of waves. It destroys coastlines and coastal settlements. It is also known as a tidal wave.

So, the correct option is (b) Tsunami.

6 0
2 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
Acceleration changes and objects...
brilliants [131]

Answer:

speed and direction

Explanation:

Acceleration is the rate of change of speed and direction.

3 0
2 years ago
How can the density of irregularly shaped objects can be calculated?
Aleks04 [339]
1.Use the balance to find the mass of the object. Record the value on the "Density Data Chart."
2.Pour water into a graduated cylinder up to an easily-read value, such as 50 milliliters and record the number.
3.Drop the object into the cylinder and record the new value in millimeters.
4.The difference between the two numbers is the object's volume. Remember that 1 milliliter is equal to 1 cubic centimeter. Record the volume on the data chart.
5.Compute the density of the object by dividing the mass value by the volume value. Record the density on the data chart.
6 0
3 years ago
Consider 3.5 kg of austenite containing 0.95 wt% c and cooled to below 727°c (1341°f). (a) what is the proeutectoid phase? (b) h
vladimir2022 [97]
A. The proeutectoid phase is Fe₃c because 0.95 wt/c  is greater than the eutectoid composition which is 0.76 wt/c

b.  We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is 
(0.14) × (3.5kg) = 0.49kg

c.  We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to 
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.
4 0
3 years ago
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