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Bess [88]
3 years ago
6

A box of unknown weight is 1 m away from the fulcrum. To balance the sides, a student applies a force of 4 N at a distance of 3

m away from the fulcrum. How much force is the box exerting on the balance beam? In other words, what is the weight of the box?
This is due by 4:10
Physics
1 answer:
Solnce55 [7]3 years ago
3 0

Answer: I think A

Explanation:

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You are trying to measure the mass of several different objects when you realize that there is a large wad of gum stuck to the u
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Answer:

accuracy

Explanation:

You are trying to measure the mass of several different objects when you realize that there is a large wad of gum stuck to the underside of the balance pan. Removing the gum will improve the <u>accuracy</u> of your measurements.

6 0
2 years ago
The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr
Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by  the following equation:

dsin\theta_{n}=n\lambda   (1)

Where:

d is the width of the slit

\lambda  is the wavelength of the light

n is an integer different from zero.

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda   (2)

Now we have to find the value of \theta_{1}:

sin\theta_{1}=\frac{\lambda}{d}  

\theta_{1}=arcsin(\frac{\lambda}{d})   (3)

We know:

\lambda=104nm=104(10)^{-9}m

In addition we are told the diffraction grating has 5000 slits per mm, this means:

d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}

Substituting the known values in (3):

\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})

\theta_{1}=arcsin(0.52)

<u>Finally:</u>

\theta_{1}=31.33\º >>>This is the first-order diffraction angle

4 0
3 years ago
How does a savanna differ from a grassland?
Pepsi [2]

Answer:

B. Savannas have shrubs and isolated trees, while grasslands contain grasses.

8 0
3 years ago
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If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
Sphinxa [80]

The line that should be secured first in pushing the boat away from the dock in preparation to dock is the bow line. When the bow line is secured, it is best to reverse it and turn to the dock, this will engage the line to tighten in a way that will help it swing back in the dock.

4 0
3 years ago
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A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi
Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0
3 years ago
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