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crimeas [40]
2 years ago
11

What is the magnitude of the momentum of a 11kg object moving at 2.2 m/s?

Physics
1 answer:
tangare [24]2 years ago
4 0

Answer:

<h3>The answer is 24.2 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 11 kg

velocity = 2.2 m/s

We have

momentum = 11 × 2.2

We have the final answer as

<h3>24.2 kgm/s</h3>

Hope this helps you

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A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from
fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

4 0
2 years ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
slamgirl [31]

Answer:

The answer is 3.

Explanation:

The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.

So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

I hope this answer helps.

4 0
2 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
2 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
2 years ago
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