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3 years ago

LetterA., please see attachment

Physics

2 answers:

Anika [276]3 years ago

7 0

The air of atmosphere contains oxygen which helps us in breathing and digestion and respiration of food
The nitrogen in air is helpful for plants and fertilizers
The humidity or water in atmosphere helps in rain

8090 [49]3 years ago

3 0

Rain causing streams to flood would be an
example of an interaction between which two
spheres? Hydrosphere and atmosphere.

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Explain how there can be more heat energy in an iceberg than in a pot of boiling water

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The tension of cold icebergs rubbing creates heat and it can also be how large the quantity is

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4 years ago

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A circuit is shown in the picture above. The chemical energy in the battery is used to light up the light bulb on the other end.

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To be honest, the picture is so far above that I can't see it at all.
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4 years ago

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If a car moves with a speed of 50 km / h and thirty minutes later it has a speed of 30 km / h, it indicates that the car is slow

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The car has slowed down because of the deceleration or decelerating force acting on it. The decelerating force acting on it is generated by the application of brakes or the frictional force acting in the opposite direction of car's motion.

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3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

lesya [120]

Answer:

Impulse in the first collision is greater

Explanation:

u = Initial velocity of cart

v = Final velocity of cart

m = Mass of of cart

First collision

Impulse

$J=m(v-u)\\\Rightarrow J=m(0.3-(-0.3))\\\Rightarrow J=m(0.6)$

Second collision

Impulse

$J=m(v-u)\\\Rightarrow J=m(0-(-0.3))\\\Rightarrow J=m(0.3)$

Hence, the impulse in the first collision is greater than the second impact

7 0

4 years ago