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olga nikolaevna [1]
2 years ago
15

The motorcycle engine on a Kawasaki Ninja 1000 has a displacement of 1043 cubic-centimeters (cm3). In order to calculate its eng

ine displacement in cubic-inches (in3) what unit conversion factor would you use to multiply the given displacement?
A. 1 in3 / 16.4 cm3
B. 2.54 cm3 / 1 in3
C. 1 in3 / 2.54 cm3
D. 6.45 cm3 / 1 in3
Physics
1 answer:
Archy [21]2 years ago
6 0

Answer:

A. 1 in³ / 16.4 cm³

Explanation:

Inches is an imperial unit while centimeter is a metric unit.

1 inch = 2.54 cm

⇒1 inch³ = 2.54³

⇒1 inch³ = 2.54×2.54×2.54

⇒1 inch³ = 16.38 cm³ = 16.38 cc

So,

1\ cc=\frac{1}{16.38}\ inch^{3}

1043\ cc=1043 \frac{1\ inch^3}{16.38\ cm^3}\\\Rightarrow 1043\ cc= 63.67\ inch^3

So,  1 in³ / 16.4 cm³ conversion factor should be used.

You might be interested in
Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp
MatroZZZ [7]

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

6 0
2 years 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

7 0
2 years ago
The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them
hammer [34]

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

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4 0
2 years ago
A 58.7-g sample of nickel (s = 0.443 J/(g ∙ °C)), initially at 147.4°C, is placed in an insulated vessel containing 184.0 g of w
kari74 [83]

Answer:

Explanation:

Let t be the final temp

temp. drop of nickel=147.4-t

temp. gain by water=t-29.9

in equilibrium

Heat lost=Heat gained

58.7×0.443×(147.4-t)=184×4.18×(t-29.9)

58.7×0.443×147.4-58.7×0.443×t=184×4.18×t-184×4.18×29.9

     3833.00434 -26.0041 t=769.12t-22996.688

(769.12 +26.0041)t= ?

find t

3 0
2 years ago
Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li
lutik1710 [3]

Answer:

Part a)

E = 3.66 eV

Part b)

\lambda = 508.5 nm

Explanation:

Part a)

change in the energy due to decay of photon is given as

E = h\nu

here we know that

\nu = 8.88 \times 10^{14} Hz

now we have

E = (6.6 \times 10^{-34})(8.88 \times 10^{14})

E = 5.86 \times 10^{-19} J

E = 3.66 eV

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

\Delta E = \frac{2}{3}(3.66 eV)

\Delta E = 2.44 eV

now to find the wavelength we have

\Delta E = \frac{hc}{\lambda}

2.44 = \frac{1242}{\lambda}

\lambda = 508.5 nm

3 0
2 years ago
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