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STALIN [3.7K]
3 years ago
11

Two atoms that are isotopes of each other must have the same number of what?

Physics
1 answer:
Naddika [18.5K]3 years ago
6 0
<span>Option C, are those that have the same number of protons in the nucleus. The atoms isotopes are atom that belongs to the same element, they are inside the periodic table and its nucleus has a different quantity of neutrons.</span>
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Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em>

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3 years ago
What do group 2 elements have in common
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3 years ago
Select the correct exercise category for each of the specific ways to increase intensity
Alecsey [184]
The answer is cardiovascular.
3 0
3 years ago
What is the period if the block’s mass is doubled? note that you do not know the value of either m or k , so do not assume any p
Oksanka [162]

The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.

The time period T of the block with mass M attached to a spring of spring constant K is given by,

T = 2π(√M/K).

Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,

T' = 2π(√2M/K)

Putting T = 2π(√M/K) above,

T' =√2T

So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.

To know more about time period of mass, visit,

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5 0
10 months ago
A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

3 0
3 years ago
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