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3 years ago

13

Please help! Calculate velocity. Show all work!

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

v = 23.66 m/s

Explanation:

recall that one of the equations of motion may be expressed:

v² = u² + 2as,

Where

v = final velocity (we are asked to find this)

u = initial velocity = 0 m/s since we are told that it starts from rest

a = acceleration = 0.56m/s²

s = distance traveled = given as 500m

Simply substitute the known values into the equation:

v² = u² + 2as

v² = 0 + 2(0.56)(500)

v² = 560

v = √560

v = 23.66 m/s

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An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal

german

Answer:

$R = 2162 m$

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

$F_y = mg$

$F_x = \frac{mv^2}{R}$

now we know that

$F_y = F cos40$

$F_x = F sin40$

also we know that

$v = 480 km/h$

$v = 133.3 m/s$

now plug in all data in above equations

$tan 40 = \frac{v^2}{Rg}$

$R = \frac{v^2}{g tan40}$

$R = \frac{133.3^2}{9.8 tan40}$

$R = 2162 m$

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3 years ago

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron

Inessa05 [86]

Answer:

providing energy to an atom can allow the electron in its non valence shell to obtain energy and move to a higher energy orbital and act as a valence electron.

Explanation:

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3 years ago

Read 2 more answers

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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3 years ago

Jerry the mouse is running along a straight desert road at a constant velocity of 18 m/s. If a certain Tom cat wants to capture

Kruka [31]

Answer:

a) t = 1.75 s

b) x = 31.5 m

Explanation:

a) The time at which Tom should drop the net can be found using the following equation:

$y_{f} = y_{0} + v_{oy}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height = 0

y₀: is the initial height = 15 m

g: is the gravity = 9.81 m/s²

$v_{0y}$ : is the initial vertical velocity of the net = 0 (it is dropped from rest)

$0 = 15m - \frac{1}{2}9.81 m/s^{2}*t^{2}$

$t = \sqrt{\frac{2*15 m}{9.81 m/s^{2}}} = 1.75 s$

Hence, Tom should drop the net at 1.75 s before Jerry is under the bridge.

b) We can find the distance at which is Jerry when Tom drops the net as follows:

$v = \frac{x}{t}$

$x = v*t = 18 m/s*1.75 m = 31.5 m$

Then, Jerry is at 31.5 meters from the bridge when Jerry drops the net.

I hope it helps you!

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2 years ago

In " m a x f o r t e d u c a t i o n " . what is probability of vowels.

LiRa [457]

Answer:

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2 years ago