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Charra [1.4K]
3 years ago
14

1.A motorcycle’s velocity at the top of the hill is 11.0 m/s. 4.0 seconds later it reaches the bottom of the hill with a velocit

y of 20.0 m/s. What is the acceleration of the motorcycle?
2.A motorcycle’s velocity at the top of the hill is 11.0 m/s. 4.0 seconds later it reaches the bottom of the hill with a velocity of 20.0 m/s. What is the acceleration of the motorcycle?
Physics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

1) as far as I remember

Let's take 20 as vf (final velocity) and 11 as (initial velocity) and 4 as time

So we would use this formula a=vf-vi/t

So 20-11/4

Asnwer 2.25

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Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve
Stella [2.4K]

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

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3 years ago
Severus Snape knows that density of his powder is 3.00 g/cm3 . He also knows he needs 3.00 cm3 of this powder. What mass in gram
harina [27]

Answer:

He requires 1 gram of mass.

Explanation:

The density is defined as:

\rho = \frac{m}{V}  (1)

Where m is the mass and V is the volume.

Then, m can be isolated from equation 1 in order to determine the mass.

m = \rho \cdot V (2)

m = (3.00g/cm^{3})(3.00cm^{3})  

m = 1g  

Hence, he requires 1 gram of mass.

8 0
3 years ago
Maddie wanted to make a sugar-water solution to put in her hummingbird feeder. She found some sugar cubes and was about to drop
NNADVOKAT [17]

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this is a good suggestion

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3 years ago
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An object in free fall is dropped from a building. Its starting velocity is 0 m/s. Ignoring the effects of air resistance, what
Amanda [17]

Answer:

30m/s^2

Explanation:

Acceleration=Final Velocity-Initial Velocity/Time

10m/s^2= Final Velocity-0m/s/3

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7 0
3 years ago
An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a
Zolol [24]

Answer:

a) \omega = 0.421\,\frac{rev}{s}, b) \Delta \theta = 0.066\,rev, c) v = 0.966\,\frac{mm}{s}, d) a = 3.293\,\frac{mm}{s^{2}}

Explanation:

a) The angular velocity of the turntable after 0.200\,s.

\omega = \omega_{o} + \alpha\cdot \Delta t

\omega = 0.240\,\frac{rev}{s}  + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)

\omega = 0.421\,\frac{rev}{s}

b) The change in angular position is:

\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot  \alpha \cdot t^{2}

\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}

\Delta \theta = 0.066\,rev

c) The tangential speed of a point on the rim of the turn-table:

v = r\cdot \omega

v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

v = 9.655\times 10^{-4}\,\frac{m}{s}

v = 0.966\,\frac{mm}{s}

d) The tangential and normal components of the acceleration of the turn-table:

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a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}

a_{t} = 2.078\,\frac{mm}{s}

a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}

a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

a_{n} = 2.554\,\frac{mm}{s^{2}}

The magnitude of the resultant acceleration is:

a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}

a = 3.293\,\frac{mm}{s^{2}}

8 0
3 years ago
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