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klio [65]
2 years ago
5

How far could a rabbit run if it ran 36km/h for 5.0min?

Physics
2 answers:
krok68 [10]2 years ago
5 0

Distance = (speed) x (time)

The question GIVES us both the speed and the time.  Sadly, though, in order to work with them, we need them to both have the same units, but the information we're given has two different units for time:  

The speed is in (km per HOUR), but the time is in MINUTES.  We'll have to change one or the other before we can multiply them.

Well, 5.0 minutes is (5/60) = 1/12 of an hour.  We can use that.

Distance = (36 km/hr) x (1/12 hour)

Distance = (36/12) (km-hr/hr)

<em>Distance = 3 km . </em>

<em> = = = = = = = = = = </em>

There's a slightly different way to handle it.  If the problem was more complicated, this way would actually turn out to be a better way, and you'd have LESS chance of making a mistake if you did it this way:

Distance = (36 km/hr) x (5.0 min)

Distance = (36 x 5)  km - min / hr

Distance = 180 km · min / hr

This is a correct answer, but the unit is awkward.  We can change the unit at THIS point in our calculation.  All we have to do is multiply the answer by ' 1 ' !

We'll find a fraction that has the same number on top or bottom, but in different units, and we'll multiply our answer by this fraction.  It won't CHANGE our answer because we're only multiplying it by ' 1 '; but If we're clever about how we build the fraction, it can solve the problem of the units for us.

The fraction to use is . . .  (1 hour / 60 minutes) .

The top and the bottom are the same quantity, so the value of the fraction is ' 1 ', and we can multiply our answer by it without changing the answer. But watch what it does to our units:

Distance = (180 km · min / hr) · (1 hr / 60 min)

Distance = (180 · 1 / 60) (km · min · hr / hr · min)

(Do some canceling, and you get ...)

<em>Distance = 3 km</em>


Thank you.  Thank you very much.  Thank you.  I'll be here until Thursday.  don't forget to tip your waiter.


Korolek [52]2 years ago
4 0

3 kilometers, it is just 5/60 or 1/12 multiplied by 36.

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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

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Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

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5 0
1 year ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

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2 years ago
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s344n2d4d5 [400]

Answer:

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m  = mass of the bat = 0.96 kg

d  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

T  = Period of oscillation = 1.35 sec

I = moment of inertia of the bat

Period of oscillation is given as

T = 2\pi \sqrt{\frac{I}{mgd}}

1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}

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2 years ago
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Maurinko [17]

Answer:

v = 0.8 m/s towards left

Explanation:

As we know that there is no external force on the system of two cart so total momentum of the system is conserved

so we will say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data into the above equation

3(1) + 5(-2) = 3(-1) + 5 v

here we assumed that left direction of motion is negative while right direction is positive

so we can solve it for speed v now

3 - 10 = - 3 + 5 v

5 v = -4

v = -0.8 m/s

3 0
2 years ago
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