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Nesterboy [21]
3 years ago
8

2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe

s it take for the football to rise to the highest point of its trajectory?
Physics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

t = \frac{2vsin \theta}{g}

t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

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A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?
Ilya [14]

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

<u>And now we have to write the temperature in kelvins.</u>

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

<u>Let us find it now.</u>

64°C + 273 = 337k

Therefore,

64°C ⇒ <u>337k</u>

8 0
1 year ago
Explain Alfred wegener's hypothesis
grandymaker [24]
He proposed the theory of Continental Drift. He believed that all of the continents were once joined together in a super continent he called Pangea (not sure if that's spelled right I studied this awhile ago). He proved this by showing similarities in fossils in completely different continents and showing how well the continents could fit together.
8 0
3 years ago
F= (9.3 x 105 )(4.2 x 10-15)
kvv77 [185]

Answer:

3.906E-9 or 3.906 x 10-9

Explanation:

8 0
3 years ago
Please look at the diagram and ans that question
TEA [102]
I can see that they are running away like my dad did
3 0
2 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
2 years ago
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