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abruzzese [7]
3 years ago
7

A light bulb has a voltage of 36 and a current of 8 A. Calculate the resistance of the light bulb

Physics
1 answer:
timurjin [86]3 years ago
4 0
R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm
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A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


6 0
3 years ago
The net force on an object moving with constant speed in circular motion is in which direction
ipn [44]

Answer:

the net force is the same direction as the acceleration

Explanation:

so toward the center of the circle about which the object is constantly moving.

7 0
2 years ago
Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)
anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

H_0 = Hubble constant = \frac{65}{3.2\times 10^6}\ ly

D = Distance to the galaxy

According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

The galaxy is 1.4 billion light years away

5 0
3 years ago
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0
3 years ago
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