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azamat
3 years ago
6

A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15 minutes, w

hat is the average speed of the bus (m/s)?
Physics
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

The average speed of the bus, v = 1.55 m/s

Explanation:

Given that,

Number of blocks traveled by bus towards east = 6

Number of blocks traveled by bus towards north = 8

Length of each block = 100 m

Distance traveled by bus towards east 6 x 100 = 600 m

Distance traveled by bus towards north 8 x 100 = 800 m

The total distance traveled, d = 600 + 800 = 1400

Time taken by the bus to travel is, t = 15 minutes

The velocity is given by the formula

                                  v = d/t  m

Substituting the values in the above equation

                                   v = 1400 m /(15 x 60) s

                                      = 1.55 m/s

Hence, the average speed of the bus, v = 1.55 m/s

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If the mass of an object increases by a factor 2, kinetic energy?
emmainna [20.7K]

Answer:

A) Increases by a factor of 2

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Given that mass, m = 2m

Substituting into the equation, we have;

K.E = ½mv²

K.E = ½*2mv²

Cross-multiplying, we have;

2K.E = 2mv²

Hence, if the mass of an object increases by a factor 2, kinetic energy is increased by a factor of 2.

5 0
3 years ago
Is it possible to calculate the displacement based on an elapsed time from a position time graph?
Nataly [62]
No I don’t think so. But it worth a try tho. Try it out.
5 0
3 years ago
In this model, the velocity of the spacecraft at position 2 is the velocity of the craft at position 4. at position 1, the direc
Free_Kalibri [48]

1. The velocity of the spacecraft at position 2 is greater than the velocity of the craft at position 4.

This is due the gravity field of the Earth is used to accelerate the craft. This is true when in a specific point the direction of the movement of the craft is the same direction of the movement of the planet.

In this case the craft will be “catched” by the Earth’s gravitational field, making the craft  to enter a circular orbit.

2. At point 1, the direction of the spacecraft changes because of the gravitational force between earth and the spacecraft.

As explained in the first answer, this is the exact point where the trajectory of the spacecraft enters into a circular orbit because of the attraction due gravity of the Earth and therefore changes its direction.

3. Position 3 represents the orbital path of Earth

Being this the orbital path of the Earth and considering the trajectory of the craft, the condition of accelerating the craft is accomplished. If the orbital path of the Earth were the opposite, the effect on the craft would be braking.

Note all of these is related to the gravitational assistance, this consists in a maneuver in which the energy of the gravitational field of a planet or satellite is used to obtain an acceleration or braking of the probe or craft, changing its trajectory.

To learn more about velocity of the spacecraft : brainly.com/question/11900446

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7 0
11 months ago
Which two statements about kinetic energy are true
gtnhenbr [62]

Answer:

A, C

Explanation:

It can be found in objects like a swing. It creates kinetic energy when it is coming back or going away, while moving.

8 0
2 years ago
How far (in meters) above the earth's surface will the acceleration of gravity be 21.0 % of what it is on the surface?
weeeeeb [17]

Answer:

7532m

Explanation:

Gravity on the surface of the earth with radius R is given by:

F_1=\frac{Gm_{earth}m_{object}}{R^2_{earth}}

Gravity a distance r above the surface:

F_2=\frac{Gm_{earth}m_{object}}{(R+r)^2}

How big is r if:

F_2=0.21\times F_1

You get the following equation:

R^2=0.21(R+r)^2=0.21R^2+0.41Rr+0.21r^2

Solve for r:

r^2+2Rr-\frac{79}{21} R^2=0

r=-R+\sqrt{R^2+\frac{79}{21}R^2}=-R+\sqrt{\frac{100}{21}R^2}=(\frac{10}{\sqrt{21}}-1)R

8 0
3 years ago
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