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3 years ago

Jill applies a force of 15 N to a wrench. If

1 answer:

6 0

Jill is the input, as she creates the force. The wrench is the output because it gives the force to the finish peace of the chain.

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A cubic meter (m³) is ______ a cubic centimeter (cm³).

valentina_108 [34]

Answer:

C. equal to

Explanation:

1 Cubic meter (m³) is equal to 1000000 cubic centimeters (cm³). To convert cubic meters to cubic centimeters, multiply the cubic meter value by 1000000.

4 0

2 years ago

In order to find the resultant of two vectors we must use the pythagoran therom, a +b2-2. Where the crepresents the resultant ve

nadezda [96]

Answer:

Furthermore, the Pythagorean theorem works when the two added vectors are at right angles to one another - such as for adding a north vector and an east vector.

8 0

2 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

A person is lifting a heavy box using a lever. What is the purpose of the lever in this situation?

ruslelena [56]

Answer:

to reduce the force needed to lift the box and change the direction of the force

Explanation:

1. "A lever consists of a rigid bar that is able to pivot at one point. This point of rotation is known as the fulcrum. A force is applied at some point away from the fulcrum (typically called the effort)."

By this definition, we know that force is needed to lift an object using a lever.

2. "When the input and output forces are on opposite sides of the fulcrum, the lever changes the direction of the applied force. This occurs only with first-class levers. When both the input and output forces are on the same side of the fulcrum, the direction of the applied force does not change"

For example, on a sew saw, if a force is applied on one end, you on the other side/end would go up, meaning a change in direction.

3. Lastly, we know a lever is typically used to reduce work, in other words, the force needed to move something.

Basically, if we were to put a lever into an equation:

reduced force + change in direction = lever

(the expection) unless load and force are on the same side, there will be no change in direction.

For example, if you and your friend sit on the same side of a sew saw, the sew saw would not go up or down, meaning no change in direction.

So if not stated otherwise you can assume the load and force are on opposite sides. The purpose of a lever in that situation would be to reduce the force needed to lift the box and change the direction of the force.

*While reading my explanation, it may be helpful to look up a diagram containing a lever, with a load, fulcrum, and applied force.

6 0

2 years ago

Examine the spectra of the four unknown substances shown below. What can you conclude?

oksian1 [2.3K]

Line spectra are obtained when individual elements are heated using a high-voltage electrical discharge. This heating causes excitation of the element and a subsequent emission of distinct lines of colored light are obtained. Each element has its own unique emission line spectrum; therefore, if any of the tested substances were the same, their spectra would match. However, this is not the case so none of the substances are the same.
hope it helps!

7 0

3 years ago

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