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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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After a collision between two different massed objects; the larger objects accelerate at a faster rate than the smaller object?
Nitella [24]

Answer: Things continue doing what they are doing unless a force is applied to it. Objects have a natural tendency to resist change. This is INERTIA. Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects, so true

Explanation:

Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop. The Cadillac has more of a tendency to stay stationary (or continue moving), and resist a change in motion than a bicycle.

6 0
3 years ago
What are dimensionless quantities??​
bearhunter [10]

Answer:

Characteristic numbers are dimensionless numbers used in fluid dynamics to describe a character of the flow. To compare a real situation with a small-scale model it is necessary to keep the important characteristic numbers the same. Names of these numbers were standardized in ISO 31, part 12.

Explanation:

7 0
3 years ago
A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s
Alisiya [41]

Answer:

v_{ic}=92.53 m/s

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

p_{i}=p_{f}

m_{c}v_{ic}=(m_{c}+m_{w})V (1)

  • m(c) is the mass of stick clay
  • m(w) is the mass of the wooden block
  • v(ic) is the initial velocity of clay
  • V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

\Delta E=W

\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d

\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd

We can find V from this equation:

V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s

Now, let's put V into the equation (1) and find v(ic)

v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s

I hope it helps you!  

<u />

8 0
3 years ago
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alekssr [168]
I think The answer is c
5 0
3 years ago
If a rock weighing 2,200 N is dropped from a height of 15 m, what is its kenetic energy just before it hits the ground
ycow [4]

Kinetic energy of the rock just before it hits the ground=KE=33000 J

Explanation:

Weight= 2200N

mg=2200

m(9.8)=2200

m=224.5 kg

initial velocity=0

final velocity =V

using kinematic equation V²=Vi²+2gh

V²=0+2 (9.8)(15)

V=17.1 m/s

now kinetic energy= 1/2 mV²

KE= 1/2 (224.5)(17.1)²

KE=33000 J

Thus the kinetic energy of the rock just before it hits the ground=33000 J

7 0
3 years ago
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