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Advocard [28]
3 years ago
6

1. A car begins at a speed of 3 m/s and accelerates at 2 m/s2

Physics
1 answer:
mina [271]3 years ago
8 0

Explanation:

v^2 = u^2 + 2as

v^2 = 9 + 160

v^2 = 169

v = 13

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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
You turn on your car's headlights while driving at night. What transformation is taking place?
Galina-37 [17]

Answer: C.

Explanation:

7 0
3 years ago
Read 2 more answers
Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t
zlopas [31]

Answer:

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

Explanation:

As we know that the wave equation is given as

y = A sin(\omega t - k x + \phi_0)

now we have

A = 0.19 m

\lambda = 2.6 m

so we have

k = \frac{2\pi}{\lambda}

k = \frac{2\pi}{2.6}

k = 2.42  per m

also we have

T = 1.2 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{1.2}

\omega = 5.23 rad/s

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

\phi_0 = \frac{\pi}{2}

so we have

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

6 0
3 years ago
How can isotopes of a given element be identified?​
marin [14]

Answer: All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom. ... Each atomic number identifies a specific element, but not the isotope; an atom of a given element may have a wide range in its number of neutrons.

Explanation: Hopefully this helps. Have a great day.

6 0
2 years ago
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is
mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= \frac{1}{2}at^2

h= \frac{1}{2}4.71\times4.96^2

h= 58.012 m

h≅ 58 m

4 0
3 years ago
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