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WARRIOR [948]
2 years ago
6

A water balloon is dropped off the top of a building and bursts when it hits

Physics
1 answer:
Dennis_Churaev [7]2 years ago
5 0

Answer:

The building is 176.4 meters high.

Explanation:

We use the formula for kinematics of an accelerated object (in this case acceleration of gravity):

Distance travelled = 1/2 g  t^2

Distance travelled = 1/2 9.8  (6)^2 = 176.4 meters

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The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux
Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

V =\pi r^2 h

Put the value into the formula

V=\pi\times(1.524)^2\times 3.048

V= 22.23\m^3

V = 22.23\times10^{3}\ Liter

We need to calculate the maximum power density in the reactor

Using formula of power density

P=\dfrac{E}{V}

Where, P = power density

E = energy

V = volume

Put the value into the formula

P=\dfrac{835\times10^{6}}{22.23\times10^{3}}

P=37561.85 = 37.562\times KW/L

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0
3 years ago
Fill in the blank: In the Northern Hemisphere, June 21 has ______________ than December 21.
Pavel [41]
Answer: I believe is A

Explanation: days are shorter in the winter
3 0
2 years ago
What is the wavelength of a light with a frequency of 6.98x10to the 14th power
NeX [460]
It is callled do it your self  you you you
5 0
3 years ago
. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42
rusak2 [61]

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position.  The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

5 0
3 years ago
1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0
3 years ago
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