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2 years ago

6

Can someone help me please

2 answers:

iogann1982 [59]2 years ago

6 0

Examine the water molecules in the animation. After the wave has passed which of the following is true?

Eva8 [605]2 years ago

4 0

Explanation:

<h3>Density </h3>

unit name : kilogram per cubic metres. ρ = m/V,

symbol : kg/m³.

<h3>Pressure </h3>

unit name : Pascal.

symbol : Pa.

<h3>Work/Energy</h3>

unit name : Joule

symbol : J

<h3>Power </h3>

unit name : Watt

symbol : W

<h3>Force </h3>

unit name : Newton

symbol : N.

please try to do the derivation on your own, I left it for you.

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A light-year measures the _______ that light travels in 1 year.

o-na [289]

<span>A light-year measures the distance that light travels in 1 year.

Answer : B ) Distance

-Hope this helps.</span>

4 0

3 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

<u>Answer:</u>

<em>Thunderbird is 995.157 meters behind the Mercedes</em>

<u>Explanation:</u>

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

$t= \frac{(v-u)}{a}$

$t= \frac{(71-0)}{6.81}= 10.425 s$

$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

= $1595.157-600=995.157 m$

Thus the Mercedes is 995.157 m ahead of the Thunderbird.

6 0

2 years ago

A car started from a rest and accelerated at 9.54 m/s2 for 6.5 seconds. How much distance was covered by the car?

ElenaW [278]

Explanation:

S =ut + 1/2at^2

S = 0×6.5 + (1/2 × 9.54) × 6.5^2

S =0 + 4.77 ×42.25

S=201.5m

5 0

2 years ago

Vinil7 [7]

Answer:

B porque tiene energía n..

3 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago