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levacccp [35]
3 years ago
15

How many planets in the solar system

Physics
2 answers:
Nataliya [291]3 years ago
7 0

Answer:

There are eight planets in our Solar System.

Explanation:

igomit [66]3 years ago
3 0

Answer: the are eight planets in the solar system

Explanation:

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune

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lesantik [10]
An ion has a negative charge
5 0
3 years ago
A ski lift has a one-way length of 1 km and a vertical rise of200m. The chairs are spaced 20 m apart, and each chair can seatthr
Strike441 [17]

Answer:

a) 68.125 KW

b) 43.04 KW

Explanation:

Distance =d= 1 km

Height = h = 200 m

Spacing between chairs =D= 20 m

No. of people per chair = 3

Speed = V= 10 km/h= 10000m\3600 s=2.8 m/s

mass of chair = 250 kg

Work to operate sky lift

W= mgh

Number of chairs any moment operational = N= 1 km/20=1000m/20=50

So, total mass of chairs = 50 × 250 =12500kg

so, w= mgh=12500×9.8×200=2452500 j

Power is rate of work - we need time for operation time of lift

Time= t = distance/speedf= 1km/(10km/h)=1km/(10 km/3600s)=360s

So, Power P= Work/time=w/t=12500/360=68125 j/s=68.125 KW

Now calculating power for operating speed in 5 sec

We calculate accelleration=a for 5 sec

a= speed / time= V/52.8/5=0.28 m/sec2

for vertical acceleration we calculate θ angle first;

tanθ= height /distance= 200/1000= 0.2

==> θ=11.34°

Vertical acceleration = a₁=a sinθ= 0.28× sin 11.34=0.10835 m/sec2

to calculate height gained during startup use;

S=vit+1/2at2

here s=H

vi=0m/s

t=5 sec

==> H = 0+1/2a₁×t=0.5×0.10835 ×5²=0.5×0.10835×5×5=1.362 m

Total Work =mgH+0.5×m×V²=12500(9.8×1.362+0.5×2.8×2.8)=215240.56 j

Again power = work / time=215240.56/5=43048.112 W=43.04 KW

6 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in
valkas [14]

Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

  • \rm g = 9.81\; N \cdot kg^{-1}.
  • \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}.
  • Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

3 0
3 years ago
Please help.....Which of the following is most likely the distance between the Milky Way and another galaxy?
bagirrra123 [75]

Answer:

The answer would be 10.8 billion light years

6 0
3 years ago
True or false? rocks have organic materials and fossil remains in them
Elina [12.6K]

Answer:

trueeeee

Explanation:

3 0
2 years ago
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