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Elina [12.6K]
3 years ago
9

A crow is flying horizontally with a constant speed of 2.45 m/s when it releases a clam from its beak. The clam lands on the roc

ky beach 2.30 s later. Consider the moment just before the clam lands. (Neglect air resistance.). . a)What is its horizontal component of velocity? (in m/s). (b) What is its vertical component of velocity? (in m/s)
Physics
1 answer:
Bezzdna [24]3 years ago
3 0
<span>Neglecting air resistance, the horizontal velocity will not have changed (i.e. it remains 2.45 m/s). The vertical velocity starts at 0 m/s and accelerates for 2.30 s (due to gravity). 9.8 = (x-0)/2.3 x = 22.54 Therefore your vertical velocity is 22.54 m/s.
</span>a) It would remain the same, because there is no force, and thus acceleration, in that direction. So, it would just be 2.45 m/s. b) Use<span>v=<span>v0</span>+at</span> Because <span>v0</span><span>=0 in the y direction, it is just v=at It said that t=2.3 s, and a=9.8. Plug that in... v=2.3*9.8=22.54 m/s</span>
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A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped from rest at some higher elevation at the ins
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Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

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Which of the following fields of science has more scientific theories and fewer scientific laws?
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Answer:

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Newton's first law of motion.

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Read 2 more answers
A stunt-rider on a motorcycle rides down a ramp and into a vertical loop-the-loop. If the diameter of the loop is 7.50 m, then t
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Answer:

v = 6.06 m/s

Explanation:

In order for the rider to pass the top of the loop without falling, his weight must be equal to the centripetal force:

Centripetal Force = Weight \\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v = \sqrt{gr}

where,

v = minimum speed of motorcycle at top of the loop = ?

g = acceleration due to gravity = 9.8 m/s²

r = radius of the loop = diameter/2 = 7.5 m/2 = 3.75 m

Therefore, using these values in equation, we get:

v = \sqrt{(9.8\ m/s^2)(3.75\ m)}

<u>v = 6.06 m/s</u>

5 0
3 years ago
Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C
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Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h =  - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

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