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3 years ago

Please help mee asappp!

Physics

1 answer:

matrenka [14]3 years ago

3 0

Answer:

Question (a)

• We shall use masses of 200g, 50g, 20g and 5g.

${ \rm{200g + 50g + 20g + 5g = 275 \: g}}$

Question (b)

• We shall use masses of 500g, 200g, 50g, 20g and 5g.

${ \rm{500g + 200g + 50g + 20g + 5g = 775 \: g}}$

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When a wave strikes a new medium, three things can happen: reflection, transmission, and absorption. Consider an echo. Do you th

kolbaska11 [484]

Answer:

YES

Explanation:

An echo may be defined as a sound which is repeated because of the sound waves that are produced are reflected back after striking a surface. Sound waves can smoothly bounce off the hard objects in the same manner as a rubber ball bounces back the ground.

When a sound wave strikes a hard surface, the sound waves gets reflected back and bounces back to the observer and produces an echo. If the sound waves strikes a soft surface it absorbs the sound.

Although the direction of a sound changes but the echo sounds in the same way as the original sound.

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4 years ago

Two point sources are vibrating in phase producing two-dimensional water wave interference. The first antinodal line on either s

Viktor [21]

For all antinodal positions we know that

$\Delta \phi = 2N\pi$

now we also know the relation between phase difference and path difference

$\Delta \phi = \frac{2\pi}{\lambda}\Delta x$

now we will have

$2N\pi = \frac{2\pi}{\lambda}\Delta x$

now from above equation we will have

$\Delta x = N\lambda$

now for the first anti node position

N = 1

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Option D is correct

4 0

3 years ago

The smallest unit of an element that has all of the properties of the element is a/an

AysviL [449]

Answer:B

Explanation: an atom is the smallest particle of an element that can take part in a chemical reaction.

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3 years ago

Your outlets at home are rated at 120 V, i.e. the two prongs have on average a potential difference of 120V. If you transfer 2.7

hodyreva [135]

Answer:

E =230.4 MJ

Explanation:

As 1 mole of electron = 6X 10^23 particles.

charge of an electron is 1.6 X 10 ^-19 C

Finding Charge:

(6X10^23 ) (2.7)(1.6X10^-19 C)

i.e. 192 K C

now to find the energy released from electrons

V=E/q

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3 years ago

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius

alisha [4.7K]

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

$B=\frac{\mu_oI}{2R}$ (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

$B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}$ (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

$\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A$

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

5 0

3 years ago

Please help mee asappp!​

Please help mee asappp!