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3 years ago

How would you make the work output of a machine greater?

1 answer:

3 0

I can think of two strategies:

1). Reduce/remove as much friction as possible from the
internal guts of the machine. Reduce the areas where two
surfaces contact each other, and use oil or grease to lubricate
the areas that can't be reduced.

2). Make the work INput greater.

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A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag

Shkiper50 [21]

The answer is 40 because you just have to do 200 divided by 5

4 0

3 years ago

If a football player has more mass they will also have more ______ ?<br> Fill in the blank

serg [7]

If a football player has more mass, they will also have more <u>momentum</u>. This is because mass is directly proportional to momentum.

3 0

3 years ago

A proton, traveling with a velocity of 4.6 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 6.0

Kobotan [32]

Answer:

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards

Explanation:

Force on moving charge in magnetic field is given as

$F = q(\vec v \times \vec B)$

here we know this is force on proton

so we have

$q = 1.6 \times 10^{-19} C$

$F = 6.0 \times 10^{-14} (-\hat j) N$

also we know that the velocity of charge is

$v = 4.6 \times 10^6 \hat i m/s$

now from above formula we have

$(6.0 \times 10^{-14}) = (1.6 \times 10^{-19})(4.6 \times 10^6)B$

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards

6 0

4 years ago

A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate

PilotLPTM [1.2K]

Answer:

$T(1)=21$

Explanation:

The equation of the position in kinematics is given:

$x(t)=x_{0}+v_{0}t+0.5at^{2}$

x(0) is the initial position, in this it is 0
v(0) is the initial velocity (20 m/s)
a is the acceleration (2 m/s²)

So the equation will be:

$x(t)=20t+0.5*2*t^{2}$

$x(t)=20t+t^{2}$

Now, the Taylor polynomial equation is:

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

$f(t)=x(t)=20t+t^{2}$

$f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t$

$f''(t)=\frac{dv(t)}{dt}=a(t)=2$

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

$f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}$

$f(0)=x(0)=0$

$f'(0)=v(0)=20$

$f''(0)=a(0)=2$

$T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}$

$T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}$

$T(t)=20t+t^{2}$

Let's put t=1 so find the how far the car moves in the next second:

$T(1)=20*1+1^{2}$

$T(1)=21$

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

4 0

3 years ago

HELP ME PLEAASSEEE ILL MARK U THE BRAINLIEST

Alika [10]

Answer:

Gravitational energy and kinetic energy

7 0

3 years ago