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2 years ago

PLEASE HELP URGETNT

Physics

1 answer:

Angelina_Jolie [31]2 years ago

6 0

Answer:

I think (d) is right answer

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Jasmine leaves her house to go for a run. She isn't paying attention to where

alukav5142 [94]

Explanation:

she ends up with a displacement of 5km north of her house and a distance of 3km

5 0

3 years ago

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

$\theta_8 = 1.12^{\circ}$ = angle to 8th max (note how m = 8 since we're comparing this to the form $\theta_m$ )

$x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}$ (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

$d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}$

Make sure your calculator is in degree mode.

-----------------

Method 2

$\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\$

-----------------

Method 3

$\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\$

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0

3 years ago

Lora (of mass 54 kg) is an expert skier. She

omeli [17]

The mechanical energy at top =Mechanical energy at bottom

Mass=m=54kg
Height=h=51m
Acceleration due to gravity=g=10m/s^2
Velocity=v=2.6m/s

$\\ \tt\longmapsto M_{initial}=M_{Final}$

Final energy at bottom=The kinetic energy

$\\ \tt\longmapsto KE=M_{initial}$

$\\ \tt\longmapsto KE=P.E_{(Top)}+K.E_{(Top)}$

$\\ \tt\longmapsto K.E=mgh+\dfrac{1}{2}mv^2$

$\\ \tt\longmapsto K.E=m\left(gh+\dfrac{v^2}{2}\right)$

$\\ \tt\longmapsto K.E=54\left((10)(51)+\dfrac{2.6^2}{2}\right)$

$\\ \tt\longmapsto K.E=54\left(510+\dfrac{6.76}{2}\right)$

$\\ \tt\longmapsto K .E=54(510+3.38)$

$\\ \tt\longmapsto K.E=54(513.38)$

$\\ \tt\longmapsto K.E=27722.52J$

$\\ \tt\longmapsto K.E=27.7KJ$

6 0

2 years ago

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0800 s, during which

Elena-2011 [213]

The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s

a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s

b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s

5 0

3 years ago

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$