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Sholpan [36]
2 years ago
8

An apple in a tree has a gravitational store of 8J. As it falls, it accelerates constantly until it hits the ground. What is the

apple's maximum kinetic energy store?
Physics
1 answer:
natita [175]2 years ago
5 0

Answer:

Explanation:

Given that on the tree the gravitational energy stored is 8J

Then, mgh = 8J.

The apple begins to fall and hit the ground, what is the maximum kinetic energy?

Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)

So,

K.E(max) = 8J

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A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current
larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

6 0
3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
2 years ago
The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele
Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 m/sec^2      

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion v^2=u^2+2as

553^2=0^2+2\times a\times 0.855

a=178835.6m/sec^2

5 0
3 years ago
Help plz I’ll mark brainliest
slavikrds [6]
Thinner at edges and its thick in the middle
3 0
2 years ago
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The period of an ocean wave is 5 seconds. What is the wave's frequency?
Oksana_A [137]

Answer:

f=1/5= 0.2 Hz

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7 0
3 years ago
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