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diamong [38]
3 years ago
15

How long did it take until all the starting material was gone?

Physics
1 answer:
Viktor [21]3 years ago
6 0
Time = (amount you started with) / (rate at which it disappeared) .
You might be interested in
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
2 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

In this case:

  • x0= 0
  • v0= 0  because the object is initially stationary
  • a= 6 \frac{m}{s^{2} }
  • t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

x=½*6 \frac{m}{s^{2} }*(15s)²

x=½*6 \frac{m}{s^{2} }*225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0
2 years ago
Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees
Vinvika [58]

Answer:

             

Explanation:

is  A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

                         

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

           

The velocity components follow the same equations we used for one-dimensional motion.

                             

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0
2 years ago
Where are globular clusters located in the milky way?.
mel-nik [20]

Answer:

Globular star clusters are located in the great spherical halo.

Explanation:

Hope this helps! :)

8 0
2 years ago
A plane is flying due east with the velocity of 90m/s. The wind is blowing out of the north at 4m/s. What is the magnitude of th
Butoxors [25]
<span> as we know that the velocity vectors are at right angles
magnitude = ?
hypotenuse of a right triangle.
v^2 = 90^2 + 4^2
v^2 = 8116
Taking the square root of both sides here we get,
v = 90.1 m/s
hope it helps
</span>
4 0
2 years ago
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