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Kazeer [188]
2 years ago
10

Help it multiple-choice 16 and 17 only

Physics
1 answer:
lilavasa [31]2 years ago
3 0

Answer:

16. c

17. d

Explanation:

16. Newton’s third law states that a force will always have an opposite but equal force as a reaction, so every force comes as a pair of action-reaction forces. For example, if you push on a book, the book also pushes on you.

17. If 1cm represents 15N, then 5cm must represent 15N*5=75N.

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
3 years ago
A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the
Whitepunk [10]

Answer:

6.93 km/h

Explanation:

To calculate her average speed, we need the "speed" formula, which is:

average speed = distance / time

You plug in your numbers and it will give you the answer.

Speed = 4.5km/0.65hr

           = 6.923 km/h

8 0
3 years ago
Read 2 more answers
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
Please help on this one?
butalik [34]

B. +Q, + W are the correct sign

7 0
3 years ago
Read 2 more answers
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