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1 year ago

As velocity__kinetic energy _ and potential energy ___

Physics

1 answer:

nataly862011 [7]1 year ago

6 0

Answer:

increases increases decreases

Explanation:

For an object falling from a height

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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose

Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0

2 years ago

Sodium and potassium are positively charged ions called cations, and chloride and phosphate form negatively charged ions called

Alecsey [184]

Sodium and potassium are positively charged ions called cations, and chloride and phosphate form negatively charged ions called anions. is true!

7 0

2 years ago

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is tr

Morgarella [4.7K]

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

$W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s$

So, the final speed of the stone as it lift the ground is 23.86 m/s.

4 0

2 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

maria [59]

Answer:

W = 0.678 rad/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

$\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh$

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I = $\frac{2}{3}mR^2$