400 N multiply by x = 100 N x 2M

Archy [21]

<span>Suppose you mixed two chemicals in the lab until you could not tell the two apart. After some time passed, a white powder formed which would not dissolve, and settled on the bottom. The mixture was first homogeneous then heterogeneous. </span>

6 0

3 years ago

Read 2 more answers

A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing

valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

4 0

4 years ago

You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

$60=10\ log\dfrac{I_1}{10^{-12}}$

10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

$\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}$

r₂=0.1 m

4 0

3 years ago

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, aft

8_murik_8 [283]

Answer:

Explanation:

Using the equation of motion v = u + at to get the speed at which the object would be travelling.

v is the final speed (in m/s)

u is the initial velocity (in m/s)

a is the acceleration (in m/s²)

t is the time taken (in secs)

Given parameters

u = 0m/s

t = 10s

a = g = 9.8m/s²

Substituting this values into the formula;

v = 0+9.8(10)

v = 0+ 98

v = 98m/s

<em>Hence the rock will be travelling at a speed of 98m/s.</em>

6 0

3 years ago

A solenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long

Airida [17]

Answer:

$0.00027646\ T$

$2.33\times 10^{-5}\ H$

-0.04194 V

Explanation:

$N_2$ = Number of turns in outer solenoid = 330

$N_1$ = Number of turns in inner solenoid = 22

$I_1$ = Current in inner solenoid = 0.14 A

$\dfrac{dI_2}{dt}$ = Rate of change of current = 1800 A/s

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius = 0.0115 m

Magnetic field is given by

$B=\mu_0\dfrac{N_2}{l}I\\\Rightarrow B=4\pi \times 10^{-7}\times \dfrac{330}{0.21}\times 0.14\\\Rightarrow B=0.00027646\ T$

The average magnetic flux through each turn of the inner solenoid is $0.00027646\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00027646\times \pi 0.0115^2\\\Rightarrow \phi=1.14862\times 10^{-7}\ wb$

Mutual inductance is given by

$M=\dfrac{N_1\phi}{I}\\\Rightarrow M=\dfrac{22\times 1.14862\times 10^{-7}}{0.14}\\\Rightarrow M=2.33\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $2.33\times 10^{-5}\ H$

Induced emf is given by

$\epsilon=-M\dfrac{dI_2}{dt}\\\Rightarrow \epsilon=-2.33\times 10^{-5}\times 1800\\\Rightarrow \epsilon=-0.04194\ V$

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

3 0

3 years ago