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3 years ago

how high does a rock rise in the air if it is thrown upward at an initial velocity 48.3ft/s and if its time of ascent is 1.5s?

Physics

1 answer:

Dima020 [189]3 years ago

6 0

here we know that rock is thrown in air with speed

$v_i = 48.3 ft/s$

here we will convert it into SI unit

$v_i = 48.3 * \frac{0.3048}{1} = 14.7 m/s$

now as the rock will rise in the air its velocity will keep on decreasing due to the gravity of earth in opposite direction

here we can say that at the highest point from the earth the speed of the rock will become zero

so we can use kinematics to find the rise of the rock in given time of 1.5 s

$h = \frac{v_f + v_i}{2}*t$

$h = \frac{0 + 14.7}{2}* 1.5$

$h = 11 m$

so the rock will rise up by total height h = 11 m

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As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

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Hi there!

We can begin by finding the acceleration of the block.

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Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

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$I\alpha = rT$

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3 years ago

A river has a steady speed 0.500 m/s. A student swims upstream a distance of1.00 km and swims back to the starting point. a) If

Ivahew [28]

Answer:

a) 33.6 min

b) 13.9 min

c) Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).

Explanation:

The problem deals with relative velocities.

Call Vr the speed of the river, which is equal to 0.500 m/s
Call Vs the speed of the student in still water, which is equal to 1.20 m/s
You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.

Now, you can state the equations for each section:

distance = speed × time
upstream: distance = (Vs - Vr) × t₁ = 1,000 m
downstream: distance = (Vs + Vr) × t₂ = 1,000 m

Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:

(Vs - Vr) × t₁ = 1,000 m
(1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
(1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
total time = t₁ + t₂ = 1429s + 588s = 2,017s
Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min

Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s

time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min

Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.

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