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3 years ago

Ice cube is dropped into a glass of water. Describe the motion of the ice cube and explain why it moves this way

Physics

2 answers:

mrs_skeptik [129]3 years ago

6 0

Gravity pulls the ice cube down when u let go of it

krek1111 [17]3 years ago

3 0

The motion of the ice cube: falling. it falls because you move it with force and drop it that makes it fall.

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If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. fi

Juliette [100K]

Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s

6 0

3 years ago

"relate the fertile phase of the menstrual cycle to the process of fertilisation"

DerKrebs [107]

Explanation:

During your menstrual cycle , Harmones make the eggs in your Ovaries mature -

• when an egg is mature , That means it's ready to be fertilized by a sperm cell .

• These hormones also make the lining of your uterus thick and spongy . So if your egg does get Fertilised , It has a nice cushy place to land and start a pregnancy .

<h3>Hope this helps </h3>

8 0

2 years ago

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coeffici

uranmaximum [27]

Answer:

V = 10.88 m/s

Explanation:

V_i =initial velocity = 0m/s

a= acceleration= gsinθ- $\mu_k$ cosθ

putting values we get

a= 9.8sin25-0.2cos25= 2.4 m/s^2

v_f= final velocity and d= displacement along the inclined plane = 10.4 m

using the equation

$v^2_f=v^2_i-2as$

$v^2_f=0^2-2(2.4)(10.4)$

v_f= 7.04 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f^2 + mgh = (0.5) m V^2

V^2 = V_f^2 + 2gh

V^2 = 7.04^2 + 2 x 9.8 x 3.5

V = 10.88 m/s

6 0

3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

2 years ago

What is the guage pressure of an object with a mass of 30 kg and 5ft height?

wariber [46]

Answer: D=pg

Explanation:

?????

8 0

2 years ago