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Sophie [7]
3 years ago
15

Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin

etic energy?
Physics
2 answers:
raketka [301]3 years ago
4 0
Velocity=3.4m/sec
Mass=30kg
so kinetic energy=1/2mv^2
=1/2×30×3.4×3.4
=15×3.4×3.4
=15×11.56
=173.4 kg m per second square

enjoy pal 
AnnZ [28]3 years ago
3 0
The formula for kinetic energy is<span>k=1/2m<span>v2</span></span>. so u can square your distance then multiply it by mass and divide it by two
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how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
Vesna [10]

2.89watts.

<h3>What is meant by sound intensity?</h3>
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  • Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.
  • Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: I\frac{P}{4\pi r^{2} }

I=1.6x10-3 w/m2

r=12m

I\frac{P}{4\pi r^{2} }

P=I4\pi r^{2}

   =(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )

   =2.89watts.

To learn more about sound intensity, refer to:

brainly.com/question/17062836

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8 0
1 year ago
5. How much time does it take for a bird flying at a speed of 45 kilometers per hour to travel a
Lyrx [107]

Answer:

40h

Explanation:

Use the velocity formula to solve

v = \frac{d}{t}

In this question, you are given velocity v = 45km/h, and you are given a distance, d = 1800km.  Time in this question is what you'll need to find.

Start by rearranging the velocity formula, to isolate for t.

v = \frac{d}{t}

Start by multiplying both sides by t

v(t) = \frac{d}{t}(t)\\\\vt = d

Then divide both sides by v.

vt\frac{1}{v} = d/v\\ \\t = \frac{d}{v}

Now that you've isolated for time, sub in your values and calculate.

t = \frac{d}{v} = \frac{1800km}{45km/h} = 40 h

8 0
2 years ago
When a negatively charged object moves in the opposite direction of an electric force field, the potential energy of the object
Simora [160]
The best and most correct answer among the choices provided by the question is decreases <span>.


</span>The potential energy of the object <span>decreases.</span>

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3 0
3 years ago
Read 2 more answers
How to find total resistance in a parallel circuit?
siniylev [52]
We can use the formula, 
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5 0
3 years ago
A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv
iren2701 [21]

Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively.  Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin.  The sphere is replaced with a hoop of the same mass and radius.  Will the value of Hmin increase, decrease, or stay the same?  Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0.  Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5.  For a hoop, k = 1.  As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above.  The track makes an angle of θ above the horizontal at point C.  Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate.  Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ.  At the maximum height, the vertical velocity is 0.  During this time, the sphere is in free fall.  The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0
2 years ago
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