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3 years ago

Which student correctly identifies a similarity and a difference between vector quantities and scalar quantities?

Physics

2 answers:

jekas [21]3 years ago

8 0

As per the question there are two physical quantities i.e scalar quantity and vector quantity.

A scalar is a physical quantity which requires only magnitude.It does not require any direction for it's complete specification.For instance we may take distance travelled,speed etc.

Unlike a scalar quantity,a vector quantity requires both magnitude as well as direction for it's complete specifications.For instance we may take displacement which is the shortest distance between two points,the velocity which is the speed in a given direction.

Hence a vector and scalar is differentiated by direction.

As per the question,the correct option will be the third option i.e C.

Carlita is right.As per her, both vector and scalar have magnitude,but only vectors have direction which is also true.

loris [4]3 years ago

8 0

C. Both scalar and vectors have magnitude. However, only vectors have direction.

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There are two types of body waves that travel out from the epicenter of an earthquake they have unique characteristics choose ea

Komok [63]

Answer:

Explanation:

distorts or shears rock as it travels through it

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3 years ago

Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you

garik1379 [7]

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

Explanation:

The gravitational attraction is given by Newton law of gravitation as follows;

$Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}$

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

M = The mass of Jupiter

m = The mass of the nearby body

R = The distance between the centers of Jupiter and the body

From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies

Therefore, as one moves away, R increases, and the strength of gravity reduces

Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.

7 0

2 years ago

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0