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2 years ago

Please help me and Can you show your work

Physics

1 answer:

RSB [31]2 years ago

4 0

Answer:

7. 1ml = 20 drops

325ml = 6500

( 325 x 20 = 6500)

there are 6500 drops in a pepsi can

8. not sure sorry ; (

9. 1 mile per hour = 88 feet per minute

45 miles per hour = 3960 feet per minute

( 88 x 45 = 3960)

there are 3960 feet per minute in 45 miles per hour

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Which type of wave affects the surface of the land by causing it to rise and fall like waves on an ocean?

denis-greek [22]

It is a seismic wave that effects the surface of the land by causing it to rise and fall like waves in an ocean. Seismic waves can be broken down into a number of different types. Specifically, it is a Raleigh surface wave that behaves much like water waves since it moves the ground up and down as it travels. Seismic waves are produced in earthquakes.

5 0

3 years ago

What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?

Contact [7]

The period is the inverse of the frequence:

$T=\dfrac{1}{f}=\dfrac{1}{440}=0.0022\overline{72}$

4 0

2 years ago

Read 2 more answers

A car with four passengers will have a shorter braking distance than a car with one braking distance. True or false. Why?

Valentin [98]

<u>Answer:</u>

The given statement is a True statement

<u>Explanation:</u>

All cars and trucks have load capacities marked on the jamb of the driver's door, as well as the owner's manual. This is very significant for braking distance.

A heavier object will require much more braking force and distance to stop, due to New ton's law of motion. “ An object in motion will tend to stay in motion “ .A heavy object will remain in motion longer and require much more force to stop. Four wheel disc brakes are great compared to front disc and rear drum configuration, in dissipating heat, and stopping more efficiently.

6 0

2 years ago

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

Read 2 more answers

50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t

r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

The potential energy of an object is given by

$PE=mgh$

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

where v is the speed of the object

When the ball is sitting on the top of the building, we have

$h=40 m$ , therefore the potential energy is not zero
$v=0$ , since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

When the ball is halfway through the fall, the height is

$h=20 m$

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

$E=PE+KE=const.$

At the top of the building,

$E=PE_{top}$

While halfway through the fall,

$PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}$

And the mechanical energy is

$E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}$

which means

$KE_{half}=\frac{E}{2}$

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

Just before the ball hits the ground, the situation is the following:

The height of the ball relative to the ground is now zero: $h=0$ . This means that the potential energy of the ball is zero: $PE=0$

The kinetic energy, instead, is not zero: in fact, the ball has gained speed during the fall, so $v\neq 0$ , and therefore the kinetic energy is not zero