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3 years ago

Is energy transformation only occurring at only point 3?

Physics

1 answer:

Paul [167]3 years ago

7 0

No, energy transformation is occurring in every point of the motion.

In fact, the ball starts from point 1 with maximum kinetic energy and zero potential energy (taking the hand of the boy as reference level). The kinetic energy converts into gravitational potential energy as it goes higher: in point 2, part of the kinetic energy has converted into potential energy (because the velocity has decreased, while the height has increased), and then when the ball reaches point 3 all the kinetic energy has converted into potential energy (because now the velocity is zero, while the height is maximum). As the ball descends (point 4), the velocity starts to increase again, therefore the kinetic energy increases and the potential energy decreases (because the height is deacreasing now).
Summarizing, energy transformation is occuring in every point of the motion.

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A 3.75 kg ball is lifted from the floor to a height of 1.5 m above the floor. What is its increase in potential energy?

inysia [295]

<span>We can use a simple equation to calculate the increase in gravitational potential energy. PE = mgh m is the mass of the object g is the acceleration due to gravity h is the change in height PE = mgh PE = (3.75 kg) (9.80 m/s^2) (1.5 m) PE = 55.1 Joules The increase in gravitational potential energy is 55.1 Joules.</span>

4 0

3 years ago

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

3 years ago

After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average

loris [4]

Answer:

0.4455 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

Total mass is

$m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg$

Here the spring constant is not given so let us assume it as $k=36000\ N/m$

Here, the forces are balanced

$mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m$

The springs are compressed by 0.4455 m

3 0

4 years ago

A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeof

weeeeeb [17]

The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>

The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
V= final velocity, U= initial velocity, S= distance and a= acceleration

<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>

Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
So, 100²-0= 2×a×1800

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ4

6 0

2 years ago

Anna35 [415]

Answer:

430.

Explanation:

If we know that 0.5 is half of a whole number, then we can simply understand that we need to 215 x 2 to get our answer.

3 0

3 years ago