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Sidana [21]
3 years ago
6

Please Help!!!!

Physics
1 answer:
oee [108]3 years ago
7 0
I think is limestone
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im postive its D) loudspeaker

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As you increase voltage, what happens to the strength of an electromagnet?
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If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?
Lynna [10]
In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
8 0
3 years ago
The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista
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I guess the problem is asking for the distance between Earth and Mars in Astronomical Units (AU).

Since 1 AU = 1.5 \cdot 10^8 km
and the distance between the two planets is
d=225 mil. km= 2.25 \cdot 10^8 km
we can convert this distance into AU by using the following proportion:
1 AU : 1.5 \cdot 10^8 km = x : 2.25 \cdot 10^8 km

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7 0
3 years ago
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A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

8 0
3 years ago
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