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3 years ago

11

Irene drops a clay ball at time t=0. The ball stops when it hits the ground. Which of the following graphs best shows the veloci

ty v of the clay ball from when it is dropped until it hits the ground?

1 answer:

deff fn [24]3 years ago

5 0

Answer:

Graph-

Explanation:

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It is used to classify the climates around the world

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A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)

a_sh-v [17]

Answer:

<em>The sprinter traveled a distance of 7.5 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

$\displaystyle a=\frac{7.5-0}{2}$

$a=3.75\ m/s^2$

Now calculate the distance:

$\displaystyle x=0*2+\frac{3.75*2^2}{2}$

$\displaystyle x=7.5\ m$

The sprinter traveled a distance of 7.5 m

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3 years ago

a cliff diver drops from rest to the water below. how many seconds does it take for the diver to go from 0 to 60. that is to go

aleksandr82 [10.1K]

U = 0, initial vertical velocity
v = 60 mph = 88 ft/s

Ignore air resistance and take g = 32 ft/s².
It t = time to attain 60 mph, then
(88 ft/s) = (32 ft/s²)*(t s)
t = 88/32 = 2.75 s

Answer: 2.75 s

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3 years ago

Four model rockets are launched in a field. The mass of each rocket and the

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Answer: it is rocket 2

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4 years ago

Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of 80.0 cm2 with

lisabon 2012 [21]

To solve this problem it is necessary to apply the concepts related to the rate of heat transfer in the materials.

By definition the rate of heat conduction transferred is understood as

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}$

Where,

$\frac{Q}{t} =$ Rate of heat transfer

k = Thermal conductivity

A = Surface area

d = Thickness

$T_{1,2} =$ Temperature at different states.

Our values are given as,

$k_w = 0.04J/s.m.\°C\\k_c = 0.84J/s.m.\°C\\A = 80*10^{-4}m^2\\d= 2cm\\T_2 = 33\°CT_1 = 10\°C$

Substituting for the wood carpet we have,

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}\\\frac{Q}{t} = \frac{(0.04)(80*10^{-4})(33-10)}{2*10^{-2}}\\\frac{Q}{t} = 0.369W$

Substituting for the ceramic we have,

$\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}\\\frac{Q}{t} = \frac{(0.84)(80*10^{-4})(33-10)}{2*10^{-2}}\\\frac{Q}{t} = 7.728W$

Therefore the rate of heat transfer for the wood carpet is 0.37W while the rate of heat transfer for the ceramic is 7.7W

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3 years ago