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Serga [27]
2 years ago
10

A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

nstant the ball is thrown. If the boy throws the ball upward at 50.0 ∘ above the horizontal, at 8.00 m/s .
How fast must the dog run to catch the ball just as it reaches the ground?
How far from the tree will the dog catch the ball?
Physics
1 answer:
ryzh [129]2 years ago
8 0

12414253

Explanation:

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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
a student weighs 1200N they are standing in an elevator that is moving downwards at a constant speed of
baherus [9]

Answer:

<em>Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?</em>

<em>This problem has been solved!</em>

<em>This problem has been solved!See the answer</em>

<em>This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of </em><em>4.9 m/s. What is the magnitude of the net force acing on the student?</em>

6 0
3 years ago
Read 2 more answers
1. A kangaroo hops 84 m to the east in 7 seconds.
DENIUS [597]

Explanation:

Given parameters:

Distance hopped  = 84m

Displacement  = 84m due east

Time  = 7s

Unknown:

Speed of kangaroo  = ?

Velocity of kangaroo  = ?

Solution:

To solve this problem,

    Speed  = \frac{distance}{time }   = \frac{84}{7}  = 12m/s

  Velocity  = \frac{displacement}{time}   = \frac{84}{7}   = 12m/s due east

3 0
3 years ago
Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tent
luda_lava [24]

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ


8 0
3 years ago
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Ablock of mass m2 on arough horinzontal surfaceis connected to aball of mass m1 by alight weight cord over alight weight frictio
aliya0001 [1]
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5 0
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