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Verizon [17]
2 years ago
11

A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.

Assuming that the extra-terrestrial’s initial speed was zero, what is the free fall acceleration of this strange world?
Physics
1 answer:
Cloud [144]2 years ago
3 0

Answer:

44.4m/s^2

Explanation:

Use the formula...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

;Acceleration of free fall = 44.4m/s^2

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A machine has an efficiency of 20 percent. Find the input work if the output work is 140 Joules
Lapatulllka [165]

Answer:

X= 700 Joules

Explanation:

The question asked about the efficiency of the work done.

The formula for efficiency is: Efficiency = (Useful output / input work) * 100%

The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.

Using the formula of efficiency,

20 = (140/X) * 100

So, we simply solve the above equation.

X= 140*100/20

X= 700 Joules

6 0
2 years ago
Read 2 more answers
You and your surfing buddy are waiting to catch a wave a few hundred meters off the beach. The waves are conveniently sinusoidal
bekas [8.4K]

Answer:

(a): The frequency of the waves is f= 0.16 Hz

Explanation:

T/4= 1.5 s

T= 6 sec

f= 1/T

f= 0.16 Hz (a)

6 0
2 years ago
Read 2 more answers
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
2 years ago
By reacting, an element that does not have a complete set of valence electrons can acquire an electron configuration similar to
Lunna [17]

Option (A) is correct.

A noble gas is different from other elements because its highest electron energy level is completely filled.The examples of noble gases are helium, neon, Argon , krypton,Xenon , radon.

All the noble gases have completely filled outermost shell. for example, Helium has two electrons and both of them are present in first shell. Neon has 10 electrons, so its electronic configuration is 2,8.It has two electrons in the first shell and eight electrons in the second shell. Thus the outermost shell of both Helium and Neon is completely filled.

This property of having completely filled outermost shells makes noble gases different from the rest of the elements.These noble gases are very less reactive .

5 0
2 years ago
Read 2 more answers
A Hooke's law bowstring is stretched x meters until a force of f newtons is applied, and then held. By what factor will the elas
Liono4ka [1.6K]

The elastic potential energy increases by a factor of 9

Explanation:

The elastic potential energy of a bowstring is given by

E=\frac{1}{2}kx^2 (1)

where

k is the spring constant

x is the elongation of the bowstring

Hooke's law states the relationship between the force applied and the elongation of an elastic object:

F=kx

where

F is the force applied

x is the elongation

We can rewrite it as

x=\frac{F}{k}

And substituting into (1),

E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}

In  this problem, the force applied to the bowstring is tripled,

F' = 3F

So the final elastic potential energy is:

E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E

So, the elastic potential energy increases by a factor of 9.

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0
2 years ago
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