How might the changes in tide affect a fishing community?

Problem: The circular blad on a radial arm saw is turning at262

kobusy [5.1K]

Answer:

Net torque, $\tau=-0.033\ N-m$

Explanation:

It is given that,

Initial angular speed of the blade, $\omega_i=262\ rad/s$

Final angular speed of the blade, $\omega_f=85\ rad/s$

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

$\tau=I\times \alpha$

The moment of inertia of the disk, $I=\dfrac{mr^2}{2}$

$\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}$

$\tau=\dfrac{0.4\times (0.13)^2}{2}\times \dfrac{85-262}{18}$

$\tau=-0.033\ N-m$

Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.

3 0

3 years ago

Perform the calculation and report your answer using sig figs. 657.70 - 26.543

Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70 - 26.543 = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0

3 years ago

A spring with a spring constant of 25.0N/m is stretched 4.50m. What is the force that the spring would apply?

alina1380 [7]

Explanation:

$from \: hookes \: law : \\ force = k \bold{.}e \\ = 25.0 \times 4.50 \\ = 112.5 \: N$

8 0

3 years ago

The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper

meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, $v_{d}$ , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

$v_{d}$ = $\frac{I}{qnA}$ ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = $\frac{I}{A}$ = $\frac{E}{p}$