Ask question

Ask question

All categories

3 years ago

15

how much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 km/hr for a dista

nce

1 answer:

NeTakaya3 years ago

8 0

I think your question is incomplete because the distance between destination and departure point isn't given in the question

You might be interested in

Which is a characteristic of centripetal acceleration?

djyliett [7]

Answer:

It is directed inward, toward the center of a circle.

5 0

3 years ago

Read 2 more answers

A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a resul

Zigmanuir [339]

Answer:

option B

Explanation:

given,

toy car is moving in circular track

speed is doubled— a change of a factor of 2

to find change in factor of acceleration

radius doesn't change

centripetal acceleration formula

= $\dfrac{v^2}{r}$

velocity is change in factor of 2

so acceleration will be change at the factor of

= $\dfrac{(2v)^2}{r}$

= 2² = 4

so the correct answer is option B

4 0

4 years ago

Please annswer will the correct letter I WILL GIVE BRAINLIEST MUST ANSWER BOTH :)

erica [24]

C and A the first one is a bit weird but thats what i got and you cant even see it that good

3 0

3 years ago

A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

geniusboy [140]

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

5 0

3 years ago

Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity v0 and glider

LenaWriter [7]

Answer:

Explanation:

Applying conservation of momentum

mA v0 + 0 = mA x 12v0 + mB x 92v0

11 mA = - 92mB .

mA / mB = 92 / 11

mA : mB = 92 : 11.

3 0

3 years ago