Can you please help me with #9?

How can an individual benefit from participating in the recommended amount of physical activity?

Orlov [11]

It will help them stay in better shape resulting in better health, being more physically active, and staying happier as a result.

3 0

3 years ago

one type of wave used to transmit a signal is a ultraviolet wave. infrared wave. radio wave. x-ray wave.

sammy [17]

Please be more specific because they all transmit signals but the most common one is the radio wave because it transmits cellphone , televition and radio signals

6 0

3 years ago

Read 2 more answers

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

3 years ago

What type of telescope is shown in Figure 24-2

lesantik [10]

Refractor, It's a refractor-esque telescope

7 0

3 years ago

Read 2 more answers

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

4 years ago