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Vikki [24]
3 years ago
7

Kuiper Belt objects are composed of which substances? Select all that apply.

Physics
1 answer:
levacccp [35]3 years ago
8 0
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron

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How to change V=72km/hr to m/s
lapo4ka [179]
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr.  They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)

=  20 meter/second
7 0
3 years ago
According to the scientific definition, which is not an example of work?
attashe74 [19]
The correct answer for the question that is being presented above is this one: "B.pushing against a car without moving it." According to the scientific definition, pushing against a car without moving it is not an example of work. Lifting a book off a desk and <span>pulling socks out of the drye are samples of work.</span>
5 0
3 years ago
With two identical light bulbs and two identical batteries how would you arrange in order to get the maximum total power to the
Norma-Jean [14]

Answer:

The batteries would be connected in series while the bulbs would be connected in parallel

Explanation:

Power (W) = VI

where V = voltage, I = current and R = resistance

from V = IR , I = V/R

Power (W) now becomes = V (V/R) = \frac{V^{2} }{R}

Power (W) =  \frac{V^{2} }{R}

from the above equation, power is directly dependent on voltage, hence the voltage has to be high for the power to be high and the power is also inversely dependent on the resistance (in this case the bulbs which act as the load)

  • We have to batteries, when batteries are connected in series the total voltage becomes the summation of the two voltages hence giving a higher voltage and when they are connected in parallel their voltage remains the same. Since we want to get higher voltage we will connect the two batteries in series.
  • we have two bulbs which are the resistance here, from the equation above the power is inversely dependent on the resistance so we would need its value to be minimal. When resistance is connected in series the resistance individual will be added to get the total resistance, hence the total resistance will be high but when the resistors are arranged in parallel you get the total resistance by applying the formula \frac{R1R2}{R1+R2} which will give us a lower resistance. Hence we would connect the bulbs in parallel.

Take note that the power from this connection should not exceed the bulbs power rating so as to avoid damage of the bulbs.

4 0
4 years ago
Question 1 (1 point)
Dmitry [639]

1). trajectory

2). person sitting in a chair

3). 490 meters

4). 65 m/s

5). False.  The projectile's displacement, velocity, and acceleration have vertical and horizontal components, but the projectile doesn't.

6). False

7). The vertical component of a projectile doesn't change due to gravity, but the vertical components of its displacement, velocity, and acceleration do.

The vertical components do NOT equal the horizontal components.

8). Decreasing if you include the effects of air resistance.  Constant if you don't.  Gravity has no effect on horizontal velocity.

9). We can't see the simulation.  But if the projectile doesn't have jets on it, then as it travels upward, its vertical velocity must decrease, because gravity is trying to not let it get away.

10). We can't see the simulation.  But if the projectile is traveling downward, we would call that "falling", and its vertical velocity must increase, because gravity is pulling it downward.

6 0
3 years ago
Read 2 more answers
Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k
Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is \bf{6.59~m/s}.

Part(B): The direction is \bf{42.7^{0}} south to east.

Explanation:

Given:

Mass of Jack, m_{1} = 59.0~Kg

Mass of Jill, m_{2} = 47..0~Kg

Initial velocity of Jack, v_{1i} = 8.00~m/s

Initial velocity of Jill, v_{2i} = 0

Final velocity of Jack, v_{1f}  5.00~m/s

The final angle made by Jack after collision, \alpha = 34.0^{0}

Consider that the final velocity of Jill be v_{2f} and it makes an angle of \beta with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}

where, v_{2f}^{x} is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s

Conservation of momentum of the system along north direction is given by

~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s

where, v_{2f}^{y} is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s

Part(B):

The direction is given by

\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}

4 0
4 years ago
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