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Vaselesa [24]
2 years ago
14

What is the resultant force on the human cannonball in the

Physics
1 answer:
Svetllana [295]2 years ago
4 0

Answer:

heart

Explanation:

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A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the e
andrew-mc [135]
Average velocity = (x( 2.08 ) - x ( 0 )) / ( 2.08 s - 0 s )
x ( 2.08 ) = 1.42 * 2.08² - 0.05 * 2.08³ =
= 1.42 * 4.3264 - 0.443456 = 6.143484 - 0.443456 ≈ 5.7 m
v = ( 5.7 m - 0 m) / (2.08 s - 0 s ) = 5.7 / 2.08 m/s = 27.4 m/s
3 0
2 years ago
TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.
horsena [70]

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

7 0
3 years ago
Read 2 more answers
When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards
aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball  = 0.0459kg

Force  = 2380N

Time taken  = 0.001s

Unknown:

Speed of the ball afterwards  = ?

Solution:

To solve this problem, we use Newton's second law of motion:

   F = m x \frac{v - u}{t}  

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

        2380  = 0.0459 x \frac{v- 0}{0.001}  

        0.0459v  = 2.38

                   v = 51.85m/s

8 0
3 years ago
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Jamie had a solid, angular crystal and noticed that it was very hard to break. Which model could represent the crystal?
VLD [36.1K]

Answer:

d

Explanation:

5 0
2 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:

a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )

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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