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meriva
3 years ago
12

What is same about all myriapods

Physics
1 answer:
seraphim [82]3 years ago
4 0
They all have segmented limbs, a hard exoskeleton, a pair of antennae and a segmented body.
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Second<br>class lever short note​
seraphim [82]

Answer:

wow

Explanation:

6 0
3 years ago
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What is the net force on this object f air = 400n (up) fgrav=600n (down)
n200080 [17]
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N
5 0
3 years ago
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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?
Olenka [21]

car starts from rest

v_i = 0

final speed attained by the car is

v_f = 5 m/s

acceleration of the car will be

a = 3.6 m/s^2

now the time to reach this final speed will be

t = \frac{v_f - v_i}{a}

t = \frac{5 - 0}{3.6}

t = 1.39 s

so it required 1.39 s to reach this final speed

6 0
3 years ago
An unknown substance is tested and found to have the following physical properties: brittle fair conductor of electricity dull (
nikklg [1K]
The answer is <span>metalloid </span>

6 0
3 years ago
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