Ask question

Ask question

All categories

2 years ago

13

A 2 kg object released from rest at the top of a tall cliff reaches a terminal speed of 37.5 m/s after it has fallen a height of

110 m . How much kinetic energy approximately did the air molecules gain from the falling object after it has fallen through this height?

1 answer:

ahrayia [7]2 years ago

3 0

Answer:

3562.25 J

Explanation:

m = mass of the object = 2 kg

v = speed of the object after falling through height of 110 m , = 37.5 m/s

h = height through which the object fall = 110 m

KE = Kinetic energy gained by air molecules

Kinetic energy gained by air molecules is given as

KE = (0.5) m v² + mgh

KE = (0.5) (2) (37.5)² + (2) (9.8) (110)

KE = 3562.25 J

You might be interested in

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

Doss [256]

Answer:

The charge to mass ratio is $-1.76\times10^{11}$

Explanation:

$Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\$

We need to find how much charge is contained in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

$\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}$

4 0

2 years ago

Read 2 more answers

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

2 years ago

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti

san4es73 [151]

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$

8 0

2 years ago

A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad

Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

$\omega$ = Angular velocity = 6 rad/s

The horizontal component of speed is given by

$v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s$

The vertical component of speed is given by

$v_v=2\ m/s$

The resultant of the two components will give us the velocity of hammer with respect to the ground

$v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s$

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

$a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2$

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0

2 years ago

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0

2 years ago

Read 2 more answers