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3 years ago

1 ) when a ball is projected upwords its time of rising is ...............the time of falling .

Physics

1 answer:

balu736 [363]3 years ago

5 0

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

y = v₀ t - ½ gt²

v₀ = 0

y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

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Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charg

zalisa [80]

Answer:

Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).

But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).

A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

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3 years ago

12.What did Chargaff said about DNA molecules?

Drupady [299]

Answer:

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3 years ago

10 PTS! HELP!

wolverine [178]

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3 years ago

Chris threw a basketball a distance of 27.5 m to score and win his

salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

$R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}$

8 0

3 years ago

A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗

shtirl [24]

Answer:

distance stop 1.52m,

velocity 4.0 m/s y^

Explanation:

The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.

a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)

Vfₓ = V₀ₓ + aₓ t

t = - V₀ₓ/aₓ

t = - 2.4/(-1.9)

t= 1.26 s

At this time the particle stops, let's find his position

X1 = V₀ₓ t + ½ aₓ t²

X1= 2.4 1.26 + ½ (-1.9) 1.26²

X1= 1.52 m

At this point the particle begins its return

b) The velocity has component x and y

As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis

Vfy= Voy + ay t

Vfy= 0 + 3.2 1.26

Vfy = 4.0 m/s

the velocity is

V = (0 x^ + 4.0 y^) m/s

c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations

X = V₀ₓ t+ ½ aₓ t²

Y = Voy t + ½ ay t²

X= 2.4 t + ½ (-1.9) t²

Y= 0 + ½ 3.2 t²

X= 2.4 t – 0.95 t²

Y= 1.6 t²

With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis

Chart to graph

Time (s) x(m) y(m)

0 0 0

0.5 0.960 0.4

1 1.45 1.6

1.50 1.46 3.6

2.00 1.00 6.4

7 0

3 years ago