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3 years ago

If an object starts at rest and moves 60 meters north along a straight line for 2 seconds, what is its average velocity?

1 answer:

8 0

Answer is d 30 m/s, North

Given: Distance d = 60 m; Time t = 2 seconds

Required: Average Velocity Vave. Vi + Vf/2

Formula: d = 1/2 at² derive acceleration a = ? and a = Vf - Vi/t

a = 2d/t²

a = 1/2(60 m)/( 2 s)²

a = 30 m/s²

Solve for final Velocity Vf = ?

a = Vf -Vi/t

Vf = at Since Initial Velocity is zero

Vf = (30 m/s²)(2 s)

Vf = 60 S

For average velocity Vave. = Vi + Vf/2

Vave. 0 + 60 m/s/2

Vave = 30 m/s North.

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Whereas the vast majority of astrophysicists support the big bang theory, many nonscientists consider the theory to be controver

lubasha [3.4K]

Answer:

Religion

Explanation:

Religion is a big factor. Many people believe that the Big Bang DID happen while some believe it to be incorrect due to religious beliefs. I can guarantee if you ask people on the street about the Big Bang, someone will say something about religion. The Theory goes against every religions beliefs as to how the universe was created.

8 0

3 years ago

A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant

Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0

3 years ago

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 10.0 m, and has vel

mihalych1998 [28]

Explanation:

Given that,

Position of the particle at t = 0,

$y=(6i+10j)\ m$

Velocity of the particle at t = 0

$u=(1i+6j)\ m$

Acceleration of the particle,

$a=(5i+7j)\ m/s^2$

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

$v=u+at$

$v=(1i+6j)+(5i+7j)10$

$v=(51i+76j)\ m/s$

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

$y'=y+ut+\dfrac{1}{2}at^2$

$y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2$

$y'=\dfrac{19}{2}i+\dfrac{39}{2}j$

(c) Magnitude of y',

$|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}$

|y'| = 21.69 meters

Direction of the y',

$tan\theta=\dfrac{y}{x}$

$tan\theta=\dfrac{39/2}{19/2}$

$\theta=64.02^{\circ}$

Hence, this is the required solution.

4 0

3 years ago

A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?

NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

Mass of cement = 62 grams.

To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

<u>Conversion:</u>

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = $\frac{62}{1000}$