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Drupady [299]
2 years ago
5

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

Physics
1 answer:
loris [4]2 years ago
4 0

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

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x = rcos</span>β
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2 years ago
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If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
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The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

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2 years ago
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A runner has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?
antiseptic1488 [7]
I could be wrong, but I'm pretty sure it's 144kg.
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A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
Planet A has mass 3M and radius R, while Planet B has mass 4M and radius 2R. They are separated by center-to-center distance 8R.
Aleksandr-060686 [28]

Answer:

Explanation:



In Newton's law of universal gravitation

F = Gm₁m₂/r²

Where G is a gravitational constant = 6.674e-11m³/kgs²

m₁ and m₂ are the masses of the two bodies or objects in question, in kilogram (kg)

r is the distance in meters between them

From the question, the rock is placed halfway between the planets

So, it's distance from planet A is 8R/2 = 4R

And it's distance from planet B is also 8R/2 = 4R

Using F = Gm₁m₂/r²

To Planet A

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet A = 3M

So Fa = G mm₂/r² = Gm(3M) / (4R)²

To Planet B,

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet B = 4M

Fb = G mm₂/r² = Gm(4M) / (4R)²

Comparing both forces together, we realise that Planet B has the largest force,

so take we F = Fb – Fa

F = Fb – Fa = Gm(4M) / (4R)² – Gm(3M) / (4R)²

F = GmM/16R²)(4–3)

F = GmM/16R²

Note that Force = Mass * Acceleration

So, F = ma

So, ma = GmM/16R² ------- Divide through by m

a = GM/16R²

From the question

M = 7.3×10^23kg

R = 5.8×10^6 m

So, a = (6.674 * 10^-11 * 7.3×10^23)/16(5.8×10^6)²

a = (48.7202 * 10^12)/16(33.64 * 10^12)

a = (48.7202 * 10^12)/(538.4 * 10^12)

a = 48.7202/538.4

a = 0.090517612960760

a = 0.091m/s² ----------Approximated

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