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2 years ago

The force created when the court pushes LeBron James upwards is equal to which force?

Physics

2 answers:

nika2105 [10]2 years ago

8 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? His force is created by the energy stored inside his muscles.

Explanation:

i hope this helps you very much

max2010maxim [7]2 years ago

7 0

Answer:

When LeBron James jumps, he is driving force into the court. How is this force created? Suggested answer: This force is created by the energy stored inside his muscles.

Explanation:

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

4 0

3 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$

which is equivalent to the gradient of the line in the velocity-time graph.

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

$a=-1 m/s^2$ is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

$T=-\frac{u}{a}=-\frac{10}{-1}=10 s$

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

$a=-1 m/s^2$ is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

$s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m$

7 0

3 years ago

A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he

Oksanka [162]

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

h = 2.49 [m]

7 0

3 years ago

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?