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Kay [80]
3 years ago
12

When potential energy ___________ , kinetic energy increases.

Physics
2 answers:
BARSIC [14]3 years ago
5 0

When potential energy <u>decreases</u>, kinetic energy increases.

Aleks04 [339]3 years ago
4 0

Answer:

I think it is decrease

correct if wrong

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What does Newton's 2nd Law explain?
kow [346]

Answer:

C. Why you must push harder to move a car farther.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Acceleration = \frac {Force}{mass}

Hence, Newton's 2nd Law explains why you must push harder to move a car farther because of its mass. Thus, it is important to increase the force that the engine provides and decrease the mass of the car.

6 0
3 years ago
Briefly explain how a resonance tube works​
Lera25 [3.4K]

Answer:

As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.

4 0
2 years ago
Read 2 more answers
B. The silica cylinder of a radiant wall heater is 0.6 m long
SIZIF [17.4K]

So,  If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

  • ε = emissivity = 1 (since we are not given),
  • σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
  • A = surface area of cylindrical wall heater = 2πrh where
  • r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
  • h = length of heater = 0.6 m, and
  • T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π  × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72  × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

6 0
3 years ago
arzan, who weighs 700 N, swings from a cliff at the end of a convenient vine thatis 20 m long. From the top of the cliff to the
vekshin1

Answer:

= 7.07 m

Explanation:

The Tarzan reaches bottom of swing after descending 2.5 m,

change in his potential energy equals his kinetic energy at bottom of swing

m g h = (1/2) m v²   ,  

hence speed v of Tarzan at bottom of swing is given as  

v = ( 2 g h )1/2

= ( 2 × 9.8 × 2.5 )1/2

= 7 m/s

At the bottom of swing, if the vine breaks, then he is moving with horizontal velocity 7 m/s in gravitational field.  

If vertical distance from ground to bottom of swing is 5 m, then time t for Tarzan to reach ground is given by

S = (1/2)g t2 or   t = (2S/g)1/2

= ( 2 × 5 / 9.8 )1/2

= 1.01 s

Horizontal distance traveled by Tarzan = 1.01 × 7

= 7.07 m

7 0
3 years ago
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
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