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3 years ago

How long does it take light to travel from jupiter to earth?.

Physics

1 answer:

valentina_108 [34]3 years ago

7 0

The answer is : 35 to 52 minutes.
Hope this helps!
Please give Brainliest!

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Can someone please help me

Zarrin [17]

They’re at a velocity of 2.0 m/s

6 0

3 years ago

How much does a 45 kg rock weigh

bazaltina [42]

Answer:

441.45N

Explanation:

Weight = mass(kg) x acceleration (m/s^2)

Weight = 45kg x 9.81m/s^2=441.45N

Recall that weight =mass of the rock x gravitational force acting on the rock

Therefore, the rock will weigh = 441.45N

7 0

3 years ago

Please help. Brainliest will be given! 25 points. Show all work.

Drupady [299]

Explanation:

Given:

v₀ₓ = 15 m/s cos 20° = 14.10 m/s

aₓ = 0 m/s²

v₀ᵧ = 15 m/s sin 20° = 5.13 m/s

aᵧ = -9.8 m/s²

t = 1.5 s

Find: Δx and Δy

Δx = v₀ₓ t + ½ aₓ t²

Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²

Δx = 21.1 m

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²

Δy = -3.33 m

4 0

4 years ago

Waves transport

Lilit [14]

Answer:

Energy only..

Explanation:

4 0

3 years ago

Read 2 more answers

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

4 years ago