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agasfer [191]
1 year ago
10

Which part of the microscope is the circular area on the stage that light passes through?

Physics
1 answer:
patriot [66]1 year ago
6 0

Answer: The part of the microscope that is the circular area is the APERTURE

I hope this helped!

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A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude
Pani-rosa [81]
The amplitude of a wave corresponds to its maximum oscillation of the wave itself. 
In our problem, the equation of the wave is
y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
y(x,t)=0.750 cm
and therefore this value corresponds to the amplitude of the wave.
4 0
3 years ago
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A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k
pychu [463]

Answer:

The magnitude of the force, B = 5 Tesla, Up (North) direction

Explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 ×10^{-5}C

F = 750 ×  -7 ×10^{-5}

F = 0.0525

But F = qvB; B = \frac{F}{qv}

where B is the magnetic field

= 0.0525 ÷ ( -7 ×10^{-5} × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).

8 0
3 years ago
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A 36-cm-circumference loop of wire is placed between the poles of an electromagnet. When a field of 0.76 T is switched on in a t
Marrrta [24]

Answer:

R = 25.3ohms

Explanation:

See attachment below.

5 0
3 years ago
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Which scientist and atomic model are correctly matched
Novosadov [1.4K]

Answer:

Rutherford and atomic model are correctly matched.

7 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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