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Arlecino [84]
3 years ago
6

What is an example of a non contact force

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

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Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?
sattari [20]
Density = Mass/Volume 
So, given mass = 20 g and volume = 40 cm^3 
By substituting in above equation, Density = 20/40 = 0.5 g/cm^3
Hope it helps.

3 0
3 years ago
What type of evidence is needed for a hypothesis to be supported or not supported?
NISA [10]

Answer:

Hypotheses must be testable, and once tested, they can be supported by evidence. If a statement is made that cannot be tested and disproved, then it is not a hypothesis.

8 0
2 years ago
9. A student walks 2 km in 30 minutes. What
Alina [70]

Answer:

4km/hr

Explanation:

There are 60 minute in an hour. If you divided 60 by 30 minutes you’d get 2. 2x 2km is 4km and 30x 2 is 60/an hour

8 0
2 years ago
A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.
Volgvan

A) 16.1 N

The magnitude of the electric force between the corks is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C is the magnitude of the charge on the first cork

q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C is the magnitude of the charge of the second cork

r = 0.12 m is the separation between the two corks

Substituting numbers into the formula, we find

F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N

B) Attractive

According to Coulomb's law, the direction of the electric force between two charged objects depends on the sign of the charge of the two objects.

In particular, we have:

- if the two objects have charges with same sign (e.g. positive-positive or negative-negative), the force is repulsive

- if the two objects have charges with opposite sign (e.g. positive-negative), the force is attractive

In this problem, we have

Cork 1 has a positive charge

Cork 2 has a negative charge

So, the force between them is attractive.

C) 2.69\cdot 10^{13}

The net charge of the negative cork is

q_2 = -4.3 \cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the negative cork is due to the presence of N excess electrons, so we can write

q_2 = Ne

and solving for N, we find the number of excess electrons:

N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}

D) 3.75\cdot 10^{13}

The net charge on the positive cork is

q_1 = +6.0\cdot 10^{-6}C

We know that the charge of a single electron is

e=-1.6\cdot 10^{-19}C

The net charge on the positive cork is due to the "absence" of N excess electrons, so we can write

q_1 = -Ne

and solving for N, we find the number of electrons lost by the cork:

N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}

6 0
3 years ago
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