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Ksivusya [100]
3 years ago
10

A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release loc

ation (A). Till its highest point (B)? Approximately how much total time is the ball in the air before it returns back to its original height (C)?
Physics
1 answer:
zvonat [6]3 years ago
4 0

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s².  The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = <em>4.08 seconds</em>. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for <em>8.16 seconds</em> up and down.

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max2010maxim [7]

Answer:

B

Explanation:

8 0
3 years ago
A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?​
harina [27]

Answer:

impulse = 8820 kg·\frac{m}{s} or 8820 N·s

Explanation:

Impulse J is equal to the average force F_{av} multiplied by the elapsed time Δt or in equation form, J = F_{av}Δt

As long as your force of 450 N is constant then that value is your average force F_{av} and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg·\frac{m}{s} or 8820 N·s.

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2 years ago
Select the correct answer.
r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

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3 0
3 years ago
Read 2 more answers
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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