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kirill115 [55]
3 years ago
7

two springs have different spring constants how could you identify the spring with the greater spring constant value

Physics
1 answer:
Liula [17]3 years ago
7 0
The answer is B, It will be stiffer
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how does the number of chromosomes in a persons sex compare with the number of chromosome in the body cell
koban [17]

no it doesn't why because I think that it is not the same but different.

4 0
3 years ago
How many planets are made up of gas
Vadim26 [7]

Answer:

The four gas giants in our solar system are Neptune, Uranus, Saturn, and Jupiter. These are also called the Jovian planets. "Jovian planet" refers to the Roman god Jupiter and was intended to indicate that all of these planets were similar to Jupiter.

Explanation:

i hope this helps

7 0
2 years ago
In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
3 years ago
An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat
Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, v_{max} = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max}  \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max}  \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s

Therefore, the maximum speed is 21.39 m/s.

3 0
3 years ago
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