In fiction, a symbol can be described as _________

PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.

AleksAgata [21]

22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.

24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.

I hope this helps!!

5 0

3 years ago

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0

4 years ago

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

Butoxors [25]

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

$V_{E} = \frac{K*q}{r}$

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

$V_{1} + V_{2} = 0$

$K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0$ (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge

By replacing r₁ = 1 - r₂ into equation (1) we have:

$K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0$ (2)

By solving equation (2) for r₂:

$r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m$

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!

8 0

3 years ago

Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is

Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is $I = 0.822I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$

The distance from the screen is $D = 3.00m$

The intensity at the central maximum is $I_o$

The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

$I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

$sin \theta = \frac{z}{D}$

if the angle between the the light at z and the central maximum is small

Then $sin \theta = \theta$

Which implies that

$\theta = \frac{z}{D}$

substituting this into the equation for the intensity

$I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

We have that

$I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$

$=I_o [\frac{sin(0.760)}{0.760}] ^2$

$I = 0.822I_o$

4 0

3 years ago

IS

jarptica [38.1K]

Answer:

5.03

Explanation:

trust me

7 0

3 years ago

In fiction, a symbol can be described as __________.