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3 years ago

Has an atmosphere of hydrogen, helium and methane, as well as storm spots

Physics

2 answers:

Jet001 [13]3 years ago

7 0

A. Jupiter. Jupiter has all of these traits.

tatuchka [14]3 years ago

3 0

C: Jupiter

Hope I helped!

Let me know if you need anything else!

~ Zoe

You might be interested in

Please help!

Trava [24]

Answer:

3 N to the right

Explanation:

There are two forces acting on the car:

- A force of 10 N towards the right

- A force of 7 N towards the left

Therefore, the net force is given by the difference between the two, since they are in opposite directions:

$F=10 N-7 N=3 N$

And the direction is to the right, since the force to the right has greater magnitude than the force to the left.

7 0

3 years ago

Read 2 more answers

A typical adult human has a mass of about 70 kg.

Misha Larkins [42]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. FM GmMmr2
= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
= 2.40 x 10-3 N

b. FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%

3 0

3 years ago

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

4 0

3 years ago

Two ions with masses of 4.39×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

Ilia_Sergeevich [38]

Answer:

7.2 cm

Explanation:

magnetic field, B = 0.301 T

speed, v = 7.92 x 10^5 m/s

mass, m = 4.39 x 10^-27 kg

q = 1.6 x 10^-19 C

The radius of singly changed ion is given by

$r = \frac{mv}{Bq}$

where, m is the mass of ion, v be the speed of ion, B is the magnetic field and q be the charge

$r = \frac{4.39\times 10^{-27}\times 7.92 \times 10^{5}}{0.301\times 1.6\times 10^{-19}}$

r = 0.072 m

r = 7.2 cm

5 0

3 years ago

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant

Vladimir79 [104]

Answer:

The magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction $(v_j)$ = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0

3 years ago