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3 years ago

PLEASE HELP!!

Physics

2 answers:

erastova [34]3 years ago

7 0

2, 3, and 4
that should be right

stich3 [128]3 years ago

5 0

1,2,4 hope this helps.

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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 baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air

Mariulka [41]

Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0

4 years ago

According to x-ray observations, the space between galaxies in a galaxy cluster is?

crimeas [40]

According to x-ray observations, the space between galaxies in a galaxy cluster is very hot. It is because the matter between galaxies (often called the intergalactic medium) is mostly hot, ionized hydrogen with bits of heavier elements such as carbon, oxygen and silicon thrown in.

Massive structures are collapsing than at earlier times. Large collapsing structures lead to higher velocity intergalactic shocks and, as a result, significant intergalactic shock heating, with some gas heated well above the $10^{4}$ K temperatures.

Heating also occurs as galaxies expel out most of the gas that fell into them. The final product is a warm/hot phase, with temperatures of > $10^{5}$ K.

Now, Let's know how do you use X-rays to make space observations?

X-radiation is absorbed by the Earth's atmosphere, so instruments to detect X-rays must be taken to high altitude by balloons, sounding rockets, and satellites.

To learn more about Galaxy Cluster, here

brainly.com/question/16557484

#SPJ4

6 0

2 years ago

A car generates 2,552 N and weighs 2,250 pounds. what is the rate of acceleration

Lilit [14]

Ans: a = 2.50 m/s^2

Explanation:
First convert the mass in its standard unit i.e. kilogram(kg):
2250 lbs = 1020.583kg

Next use Newton's Second law:
F = ma

Where F = 2552N
m = 1020.583kg

=> a = (2552/1020.583)
a = 2.50 m/s^2

6 0

3 years ago

What’s the awnser to 1 and 2

slava [35]

Answer: for 1 is number 1

and for 2 is 3

Explanation:

7 0

3 years ago