Find the measure of angle x in the figure below:

Diana would like to observe the movements of a paramecium. Which of the following tools would be best for Diana to use? A. hand

ra1l [238]

Out of the choices given, the tool that would be best for Diana to use is the light microscope. The correct answer is B.

8 0

2 years ago

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Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w

leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0

2 years ago

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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of

SCORPION-xisa [38]

Answer:

a).β=0.53 $x10^{-3}$ T

a).β=0.40 $x10^{-4}$ T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

$\beta =\frac{u_{o}*I*N}{2\pi*r}$

a).

$N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A} \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T$

b).

The distance is the radius add the cross section so:

$r_{1}=15cm+5cm\\r_{1}=20cm$

$r_{1} =20cm*\frac{1m}{100cm}=0.20m$

$\beta =\frac{u_{o}*I*N}{2\pi*r1}$

$\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T$

3 0

2 years ago

Trong mặt phẳng Oxy, chất điểm chuyển động với phương trình<br><br> Tìm quĩ đạo của chất điểm.

QveST [7]

Xin lỗi nhưng tôi cần ảnh để trả lời đúng câu hỏi của bạn, thưa ông.

Nếu bạn đăng lại với hình ảnh, tôi sẽ sẵn lòng giúp đỡ!

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6 0

2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

2 years ago