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ryzh [129]
3 years ago
13

In a given circuit, the supplied voltage is 1.5 volts. One resistor is 3 ohms and the other is 2 ohms. Use Ohm's law to determin

e the current in the circuit. 0.3 amps 1.5 amps 7.5 amps 3.3 amps
Physics
1 answer:
tino4ka555 [31]3 years ago
4 0
I = E / R

If the resistors are in series, the current is 0.3 Amperes.

If the resistors are in parallel, the current is 1.25 Amperes.
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Why plate tectonics is a scientific theory and not a scientific law.
ser-zykov [4K]

Answer:

We have not drilled to the center of the earth.

Explanation:

7 0
3 years ago
The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le
Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

3 0
3 years ago
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
During physical activity of high intensity, your heart will beat at a percentage of its maximum rate. Whi
Maurinko [17]

Answer:

D

Explanation:

If it is of very high intensity it will be 85-100%

8 0
3 years ago
The chart shows data for a moving object.
Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0
3 years ago
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