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3 years ago

As water changes state,the mass of the sample does not change.the size of the water molecules_______

Physics

1 answer:

Mamont248 [21]3 years ago

7 0

Answer:

does not change

Explanation:

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A force of 1.5 × 102 N is exerted on a charge of 1.4 × 10–7 C that is traveling at an angle of 75° to a magnetic field.

Marysya12 [62]

Option (B) is correct.The magnetic field strength=8.5 x 10² T

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge=1.4 x 10⁻⁷ C

v= velocity= 1.3 x 10⁶ m/s

B= magnetic field strength

Fm= magnetic force= 1.5 x 10² N

θ=75°

so 1.5 x 10²=(1.4 x 10⁻⁷) (1.3 x 10⁶ ) (B) sin75

B=8.5 x 10² T

6 0

3 years ago

Read 2 more answers

What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure

Ivahew [28]

<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

Pressure = Force/Area

★ Putting the values in the above formula,we get:

$\sf \rightarrow \: pressure = \dfrac{4000}{50}$

$\sf \rightarrow pressure = 80 \: N {m}^{ - 2}$

7 0

3 years ago

An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the

bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N + N--->N + N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

$L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda$

The frequency is calculated as follows;

$F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz$

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0

3 years ago

How much voltage is in the primary coil if there are 3200 windings in the

Lesechka [4]

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

$\frac{V_1}{V_2} = \frac{N_1}{N_2}$

here we know that

$V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V$

$N_1 = 500$

$N_2 = 3200$

Now we have

$\frac{V_1}{25} = \frac{500}{3200}$

$V_1 = 3.91 V$

8 0

3 years ago

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$

$A =\pi (0.2)^2$

$A =0.1256 m^2$

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

$\mu=IA$

$\mu=2(0.1256)$

$\mu=0.25 A\cdot m^2$

Therefore the magnetic dipole moment is $0.25A\cdot m^2$

6 0

3 years ago