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melomori [17]
3 years ago
7

The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6

8 cm/s. What are (a) vavg, (b) vrms, and (c) vP?
Physics
1 answer:
Leona [35]3 years ago
8 0

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s

Average velocity is 3.93 cm/s

v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s

Root mean square velocity is 4.79 cm/s

v_p=7.68-1.4=6.28\ cm/s

Velocity peak to peak is 6.28 cm/s

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goldenfox [79]

Answer:

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3 0
2 years ago
Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4
DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4,  the slope of end a of the cantilevered beam  is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the   is mathematically given as

A=\frac{PL^2}{2EI}+\frac{ML}{EI}

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion,  the slope is

A=0.0048rads

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5 0
2 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
2 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
2 years ago
The power in an electrical circuit is given by the equation P= RR, where /is
Rainbow [258]

Answer:D

Explanation:Electric power=I*I*R

=12*12*100

=14400watts

6 0
3 years ago
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