All categories

2 years ago

A 1 kg coconut falls vertically from a height of 10 meters. Use the principle of conservation of mechanical

Physics

1 answer:

Kay [80]2 years ago

5 0

Answer:

Explanation:

Question 1:

Mass=1kg

Acceleration due to gravity=9.8m/s^2

Height=10m

on the before falling it has potential energy

Potential energy=mass x acceleration due to gravity x height

Potential energy=1 x 9.8 x 10

Potential energy=98 joules

Question 2:

Potential energy=kinetic energy base base on energy transformation

Kinetic energy=(mass x (velocity)^2)➗2

98=(1 x(velocity))^2 ➗ 2

Cross multiplying

98 x 2=(velocity)^2

196=(velocity)^2

Velocity=√(196)

Velocity=14

Velocity=14m/s

You might be interested in

A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

8 0

1 year ago

During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

1 year ago

Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia

masha68 [24]

Infiltration

Explanation:

The component of the hydrologic cycle affected by impervious building such as concrete and asphalt is infiltration.

Water infiltration is a major component of the hydrologic cycle.
Concretes and other materials can prevent water from going down into the earth.
This affects the ground water system in place.
It leads to increase in surface run off and might cause inundation of an area.
Infiltration is a very important component of water cycle.
It takes water to plant root and recharges groundwater systems.
Impervious structures takes this capability away.

learn more:

Biogeochemical cycle brainly.com/question/3509510

#learnwithBrainly

3 0

2 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

1 year ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Ksenya-84 [330]

Answer:

The answer is " $6\ \frac{m}{s}$ "