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4 years ago

What would be the speed of an object just before hitting the ground if dropped 100 meters?

Physics

1 answer:

PolarNik [594]4 years ago

8 0

<u>We are given:</u>

the initial height of the object (h) = 100 m

initial velocity (u) = 0 m/s

we will let the value of g = 10 m/s/s

<u>Speed of the object just before hitting the ground:</u>

From the third equation of motion:

v² - u² = 2ah (where v is the final velocity)

replacing the variables, we get:

v² - (0)² = 2(10)(100)

v² = 2000

v = 10√20 = 44.7 m/s

Therefore, the speed of the object just before hitting the ground is 44.7 m/s

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(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c

andreyandreev [35.5K]

Answer:

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

Explanation:

The formula for the surface of the circle is:

$A(r) = \pi\cdot r^{2}$

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$

Finally, variables are replaced by known data:

$\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )$

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

4 0

4 years ago

Read 2 more answers

Calculate the mass of air of density 1.2kgm-3 in a room of dimensions 8m by 6m by 4m.

77julia77 [94]

Answer:

230.4kg

Explanation:

volume of the room = l× b×h

volume= 8×6×4

volume=192m3

density= mass/volume

hence mass= density × volume

mass= 1.2kgm-3 × 192m3

mass= 230.4kg

5 0

2 years ago

A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.50 m above. The second student cat

Mkey [24]

Answer:

a) 8.58 m/s upward

b) -2.211 m/s downward

Explanation:

Let gravitational acceleration g = -9.81m/s2. This is negative because it deceleration the upward motion of the key.

(a)We have the following equation of motion

$s = v_0t + gt^2/2$

where $v_0$ is the initial upward velocity of the keys, t = 1.1s is the time it takes for the keys to travel a distance of s = 3.5 m

$3.5 = v_01.1 - 9.81*1.1^2/2$

$3.5 = 1.1v_0 - 5.94$

$1.1v_0 = 3.5 + 5.94 = 9.44$

$v_0 = 9.44 / 1.1 = 8.58 m/s$

So the keys were thrown initially upward with a speed of 8.58 m/s

(b) If the initial velocity of the key is 8.58 m/s and it is subjected to a deceleration of 9.81m/s2 for 1.1s then the velocity right at the 1.1s instant is

$v = v_0 + gt = 8.58 - 9.81*1.1 = -2.211 m/s$

So they keys would have a downward speed of 2.211 m/s

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3 years ago

Explain what concave and convex mirrors do to parallel rays directed to their surfaces.

dezoksy [38]

Answer:

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Explanation:

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7 0

2 years ago

The container was lifted a height of 14 m

Alika [10]

Work done = Force X Distance

3 430 000J = Force X 14m

Force = 3 430 000J / 14m
= 245 000 N

Hope this helps!

8 0

2 years ago