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Elis [28]
4 years ago
10

Does anyone know question 6?

Physics
1 answer:
vfiekz [6]4 years ago
3 0

Answer:

Yes, the fuse will blow.

Explanation:

We'll begin by converting 50000mA to amperes (A) .this is illustrated below:

1000mA = 1A

Therefore, 50000mA = 50000/1000 = 50A.

From the question given above, the plug can only accommodate 5A. Now if 50000mA i.e 50A is passed through the plug, the plug will blow because the 50A is higher than what the plug can accommodate.

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Based on the your prior knowledge, how far up a hill will a Hot Wheels car (0.01 kg) go if it is traveling at 5.0 m/s?
Sloan [31]

The formula for speed id speed is equal to distance/time

so to 5.0/0.01=500m/s.

8 0
4 years ago
Plot a graph based on the passage
Mnenie [13.5K]
Hence, the horizontal velocity of the rover is 1.73 m/s
8 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
4 years ago
ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr
Free_Kalibri [48]

Answer:

\beta = 41.68°

Explanation:

according to snell's law

\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

sin\alpha = \frac{n_w}{n_g}* sin30

sin\alpha = 0.443

now applying snell's law between air and glass, so we have

\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}

sin\beta = \frac{n_g}{n_a} sin\alpha

\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]

we know that sin\alpha = 0.443

\beta = 41.68°

7 0
3 years ago
For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red
barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

z=\sqrt{\frac{c+v}{c-v}}-1

- A red dwarf moving away from Earth at 39.1 km/s :

z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}

- A blue dwarf moving away from Earth at 25.9 km/sz=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}

- A red dwarf moving transversely at 14.1 km/s

z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}

6 0
4 years ago
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