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postnew [5]
3 years ago
5

what is the resistance of a clock if it has a current of 0.30 a and runs on a 9.0-v battery? 0.033 2.7 9.3 30

Physics
2 answers:
OleMash [197]3 years ago
6 0

Answer : Resistance of a clock is  30\ \Omega.

Explanation :

It is given that :

The current flowing in the clock, I = 0.3 A

Potential difference, V = 9 V

According to Ohm's law :

V=I\times R

So,

R=\dfrac{V}{I}

R=\dfrac{9\ V}{0.3\ A}

R=30\ \Omega

\Omega is the SI unit of the resistance given by a physicist Georg Simon Ohm.

So, the correct option is (d) i.e. 30\ \Omega

Hence, this is the required solution.

Nataliya [291]3 years ago
5 0
R = V/I =  9 / 0.3

R = 30 ohms.
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Rzqust [24]

Answer: The light bends because light travels fast but it slows down in a denser medium. For example light refracts in water or it bends after passing through air. When light passes through air ( a less dense medium ) then through water ( a more dense medium ) the beam of light bends because light travels more slowly in a denser medium then it picks up its pace again once it passes. The density of the substance determines how much the light is refracted. I hope this makes sense and I hope this answered your question!! :)

4 0
3 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
B. Jerome plays middle linebacker for South's varsity football team. In a game against
weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
4 years ago
Roots that grow horizontally are called
Romashka [77]
I believe they are called Rhizomes. 
8 0
4 years ago
Read 2 more answers
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