All categories

3 years ago

What is being done about emission controls?

1 answer:

5 0

The original Clean Air Act of 1970 gave the US EPA board authority to regulate motor vehicle pollution and the agencies emission control policies and requirements have become progressively more stringent since then

You might be interested in

a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

ludmilkaskok [199]

Answer:

Position at t= 4 seconds is 144 m

Explanation:

It is given that acceleration, a = 18 t, where t is the time.

We know that Velocity, $v = \int { a} \, dt$

Substituting value of a,

Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

We know that at t = 0, v = -12 m/s

So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

We also know that displacement, $x = \int { v} \, dt$

Substituting value of v,

Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

We know that at t = 0, particle is at origin, x =0.

So, $0=3*0^3-12*0+c\\ \\ c=0$

Displacement, $x = 3t^3-12t$

At t = 4 seconds

$x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m

4 0

3 years ago

How do mountain glaciers and continental glaciers differ in terms of dimensions, thickness and patterns of movement?

tester [92]

Question:
How do mountain glaciers and continental glaciers differ in terms of dimensions, thickness and patterns of movement?

Answer:
<span>
Continental glaciers are thicker, much more expansive sheets. Mountain glaciers flow downhill as a result of gravity acting on the mass of ice. Continental glaciers move in response to pressure from the weight of material in their thick midsections.
</span>

Hope this helped!

~Shane :}

8 0

4 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0

3 years ago

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

Kipish [7]

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

$T=2\pi \sqrt{\frac{l}{g}}$ (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

$l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m$

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

5 0

3 years ago

1) La longitud del brazo de potencia de una palanca es de 0,0035 hectómetros y la del brazo de resistencia es de 55 centímetros.

Arlecino [84]

Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

6 0

3 years ago