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3 years ago

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Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl

ems when exercising. B. Bright colors work to keep you cooler while exercising. C. High-visibility clothing is important to wear in areas with moving vehicles. D. none of the above

2 answers:

allochka39001 [22]3 years ago

6 0

Answer:

C. High-visibility clothing is important to wear in areas with moving vehicles.

Explanation:

plz mark as brainliest

Mazyrski [523]3 years ago

5 0

C. High-visibility clothing is important to wear in areas with moving vehicles

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A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp

Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2$

b) Assume the grind stone is a solid disk, its moment of inertia is

$I = mR^2/2$

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

$I = 28*0.15^2/2 = 0.315 kgm^2$

So the friction torque is

$T_f = I\alpha = 0.315*0.21 = 0.06615 Nm$

The friction force is

$F_f = T_f/R = 0.06615 / 0.15 = 0.441 N$

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

$N = F_f/\mu = 0.441/0.2 = 2.205 N$

7 0

3 years ago

A student takes a ride on a ferris wheel and goes around 5 times. if the radius of the ferris wheel is 7.5 m how far did the stu

poizon [28]

If the radius is 7.5 meters, then the circumference would be 47.1239. That times five is 235.6195

4 0

3 years ago

The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal

kogti [31]

Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

3 0

2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10

loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

$\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}$

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

$\displaystyle Average \ force, \, F_{ave} \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}$

The average force exerted on the limestone floor by the droplets of water is $F_{ave}$ ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0

2 years ago