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Setler79 [48]
3 years ago
9

Saturn is moving with uniform angular speed of 2πf along the circumference of it's orbit around the sun with radius R, having ce

ntre O. At any time, the angular position of Saturn is (2πf)t and the displacement in SHM at that time t is given by x(t)=Rcos(2πf)t. Find it's acceleration.​
Physics
1 answer:
Luda [366]3 years ago
8 0

Answer:Sir haymo knows

Explanation:

he gave as homework

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An object is thrown upward with an initial speed of 18.5 m/s from a location 12.2 m above the ground. After reaching its maximum
siniylev [52]

Answer:

The speed of the object in the last instant prior to hitting the ground is -24.1 m/s

Explanation:

The equation for the position and velocity of the object will be:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the object at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

We know that at its maximum height, the velocity of the object is 0. We can obtain the time it takes the object to reach the maximum height and with that time we can calculate the maximum height:

v = v0 + g · t

0 = 18.5 m/s - 9.8 m/s² · t

-18.5 m/s / -9.8 m/s² = t

t = 1.89 s

Now,let´s find the max-height:

y = y0 + v0 · t + 1/2 · g · t²

y = 12.2 m + 18.5 m/s · 1.89 s + 1/2 ·(-9.8 m/s²) · (1.89 s)²

y = 29.7 m

Now, let´s see how much it takes the object to hit the ground:

In that instant, y = 0.

y = y0 + v0 · t + 1/2 · g · t²

0 = 29.7 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²   (notice that v0 = 0 because the object starts from its maximum height, where v = 0)

-29.7 m = -4.9 m/s² · t²

t² = -29.7 m / -4.9 m/s²

t = 2.46 s

Now, we can calculate the speed at t= 2.46 s, the instant prior to hitting the ground.

v = v0 + g · t

v = g · t

v = -9.8 m/s² · 2.46 s

v = -24.1 m/s

4 0
3 years ago
What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Ye
Alex_Xolod [135]

A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) <u>Collecting dew</u> from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

Learn more about Adaptation of cactus here brainly.com/question/12501143

#SPJ1

5 0
2 years ago
How to measure the volume of a baseball bat ( need answers ASAP )
vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

8 0
3 years ago
What is the cooling power P? Recall that the rate at which energy is removed, also called the cooling power, was described earli
gayaneshka [121]

Answer:

36.667

Explanation:

(2.2*10^4)/(10*60)

3 0
3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}

Now we use the formula

{v_f}^2-{v_i}^2=2a(x_f-x_i)

to find the distance it travels. The sled comes to a rest, so v_f=0, and let's take the starting position x_i=0 to be the origin. Then the distance traveled x_f-x_i=x_f is

-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m

6 0
3 years ago
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