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2 years ago

15

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is tr

aveling at 22.0 m/s upward. Use the work-energy theorem to find the rock's speed just as it left the ground.

1 answer:

Morgarella [4.7K]2 years ago

4 0

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

$W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s$

So, the final speed of the stone as it lift the ground is 23.86 m/s.

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A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted

Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

$F_c = F_g - F_N$

where;

Fc is the centripetal force

$F_g$ is downward force due to weight of the driver

$F_N$ is upward or normal force on the drive

$F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N$

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0

2 years ago

The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e

FrozenT [24]

Answer:

$1.1486813808\times 10^{-7}\ T$

34.46 V/m

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

c = Speed of light = $3\times 10^8\ m/s$

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

$B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T$

The magnetic field intensity is $1.1486813808\times 10^{-7}\ T$

The maximum electric field intensity is given by

$E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m$

The electric field intensity is 34.46 V/m

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2 years ago

An object moving 20 m/s

yan [13]

Answer:

<u>We are given:</u>

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

<u />

<u>Solving for Displacement:</u>

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0

2 years ago

How long does it take decaying organisms to transform into fossil fuels?

taurus [48]

Answer:

Millions or hundreds of million of years

Explanation:

It takes a very long time for decaying substances to form fossil fuels. It takes around millions or hundreds of million of years for the transformation to occur. This explains why they aren’t renewable.

Decayed plant materials usually form coal while decayed animal materials form crude oil.

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3 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

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3 years ago