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madam [21]
3 years ago
5

A frog leaps with a displacement equal to vector u and then leaps with a displacement equal to vector v, as

Physics
1 answer:
denpristay [2]3 years ago
7 0

Answer:

The answer is B on khan academy

Explanation:

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A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
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1.5 m is ur answer

Explanation:

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3 years ago
To pull a 53 kg crate across a horizontal frictionless floor, a worker applies a force of 180 N, directed 35° above the horizont
Lorico [155]

Answer

given,

mass of the crate  = 53 Kg

force applied by the worker = 180 N

Angle made with the horizontal = 35°

crate moves = 2.9 m

a) The work done equals the force in the direction of the displacement, times the displacement

W = F_x ×d

W = 180 cos 35° × 2.9

W = 427.6 J

b) A force that is perpendicular to the direction of the displacement does not do any work so work done by gravitational force is zero

c) Similarly for the normal force the work done will be zero.

d) Total work done by the crate is equal to 427.6 J

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4 years ago
Part 1: Describe the mass and surface temperature of the supergiant stars.
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The temperature of the surface is between 3,500-20,00
 supregiants would be around the top half for the y axis 
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6 0
3 years ago
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Someone please help me ?
lutik1710 [3]

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1 9 5 3 into law of inertia and 4 6 into law of acceleration and 2 7 8 into law of reaction

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How long does it take the Sun to melt a block of ice at 0∘C with a flat horizontal area 1.0 m2 and thickness 2.0 cm ? Assume tha
sineoko [7]

Answer:

It takes <em>40 hours</em> to melt the block of ice.

Explanation:

According to the principles of radiation and heat transfer respectively:

<em>ΔQ = I(dt)eAcosθ  </em>(I = Solar energy density; dt = time taken; e = emissivity; A = Area of block; θ = angle between the sun ray and the horizontal)

<em>ΔQ = mLf</em>  (ΔQ = Heat change; m = mass of ice; Lf = Specific latent heat of fusion of ice)

but m = ρV = ρ.A.<em>d</em>x, therefore, the heat transfer equation can be re-written as:

<em>ΔQ = ρ.A.dx.Lf</em>

Lets equate the radiation equation and the modified heat transfer equation, we have:

<em>ρ.A.dx.Lf = I(dt)eAcosθ</em>

<em>ρ.dx.Lf = I(dt)ecosθ    </em>(Striking out the area)

Let's make <em>dt</em> the subject of formula,

dt = ρ.dx.Lf /I.e.cosθ

ρ = Density of ice, 9.2x10^{2} Kg/m^{3}

Lf = 3.36x10^{5} J/Kg

e = 0.050

θ = 32 deg. C

Now, let's substitute the terms:

dt=\frac{(9.2x10^{2})(0.02)(3.36x10^{5} ) }{(1000)(0.050)(cos32)}

dt=14.45x10^{4} s = \frac{14.45x10^{4}}{3600} hr=40.14 hr

Therefore, the time taken for the ice to completely melt is <em>40 hours</em> (Two significant figures)

4 0
3 years ago
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