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gayaneshka [121]
3 years ago
9

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s

urface of the earth. In this problem, use 9.80 m/s2 for the acceleration due to gravity on earth. What is its mass on the earth's surface?
Physics
1 answer:
butalik [34]3 years ago
4 0

Answer:

m=5.51Kg

Explanation:

The weight of an object on Earth is given by W=mg, so we can calculate its mass by doing m=W/g, which for our values is:

m=(54N)/(9.8m/s^2)=5.51Kg

<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:

W'=ma=(5.51Kg)(1.81m/s^2)=9.97N

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EASY BRAINLIEST PLEASE HELP!!
Soloha48 [4]

Answer:

Solution given:

frequency[f]=x

velocity[V]=15000m/s

wave length=59m

we have

wave length=\frac{V}{f}

59m=\frac{15000}{x}

x=\frac{15000}{59}=254.Hz

frequency=254Hz

7 0
3 years ago
Read 2 more answers
Once a scientist has made a hypothesis, what would they typically do next? (2 points)
ipn [44]

Answer:

b-testing

Explanation:

First would be observation/research. Then the hypothesis. After that you would test your theory, conduct experiments. And finally, your conclusion- what you got from the whole process basically.

Hope this helps.

4 0
3 years ago
A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w
Goryan [66]

Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr  = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final  velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

4 0
3 years ago
A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
pav-90 [236]
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

                t² - 26.28 t = 0

                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

5 0
3 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
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