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2 years ago

6

When using a physical balance on what pan we place the object and why

1 answer:

kondaur [170]2 years ago

4 0

Answer:

wood

Explanation:

Wooden pans are usually used
Wood has less density than water so it will not sink in liquid substances derived from water.
So during physical balance it won't affect the balance which we need

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In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- ve

choli [55]

Answer:

$B_{int}=-0.015T$

Explanation:

From the question we are told that:

RF source oscillation speed $\sigma= 34 MHz$

The external field $Bext =0.78 T$ .

Pro- ton magnetic moment component $\mu=1.41 X 10-26 J/T$

Generally the equation for magnitude of $B_{int}$ is mathematically given by

$B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}$

$B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}$

$B_{int}=0.78-0.7957$

$B_{int}=-0.015T$

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3 years ago

What is kinetic energy?

Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

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3 years ago

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A dog travels 18 meters south across the backyard in 11 seconds. What is the dog's speed?

o-na [289]

The dog’s speed is
A) 0.61 m/s

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4 years ago

Which two statements best determine whether an engineering design is

pshichka [43]

Answer: A and B

Explanation:

just took the apex test

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3 years ago

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

So

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

6 0

3 years ago