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scZoUnD [109]
3 years ago
5

Which planet is least like earth? Mars,Venus, or Jupiter

Physics
2 answers:
irina1246 [14]3 years ago
8 0
Jupiter is least like Earth. Planets like Venus, and Mars are mostly rocky and are more or less similar in interbal composition. Jupiter is large, gaseous, and hydrogen-rich.
Brums [2.3K]3 years ago
7 0

Answer:

mars, reason why is because they both are diff from the size

Explanation:

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The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an
hjlf

Answer:

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

Hence the maximum height achieved by the ball is 61.22m

4 0
2 years ago
Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
Eva8 [605]

Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
  • Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude and θ.

So, for our vectors, we will have:

\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )

\vec{D}=  ( 2.121 m , -2.121 m )

and

\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )

\vec{E}= ( 2.71 m , 3.59 m )

Now, we can take the sum of the vectors

\vec{R} = \vec{D} + \vec{E}

\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )

\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m )

\vec{R} = ( 4.831 \ m , 1.469 \ m )

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}

|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}

|\vec{R}| = \sqrt{25.496 m^2}

|\vec{R}| = 5.049 m

To find the direction, we can use

\theta = arctan(\frac{R_y}{R_x})

\theta = arctan(\frac{1.469 \ m}{4.831 \ m})

\theta = arctan(0.304)

\theta = 16\°54'33''

As we are in the first quadrant, this is relative to the x axis.

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PLEWSE SOMEONE HELP ILL MARK BRAINLIST
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Answer:

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