A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft

Answers

Answer 1

Answer: I believed the answer is d

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A 600W toaster, 1200W iron and a 100W bulb are all connected to household 120V circuit. a) find the current drawn by each applianceb) the resistance of the heating element in the ironc) the total cost of running all of the appliances non-stop for 5 days if the electricity rate is $0.1 per KW-Hr.
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You have 5 cats and they each have a mass of 4kg per cat. What is the mass of all of them together?

Answers

Answer:

It would be 20kg

Explanation:

This would be just 5x4 as there are 5 cats and each are 4kg. You can also add 4, 5 times as well.

I hope Im correct

A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

$A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2$

hence,

$L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m$

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 + αΔT]

α = $4*10^(-4)^\;oC^(-1)$ for Nichrome

$R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega$

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Learn more about electrical power:

brainly.com/question/26174188

Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

Learn more: brainly.com/question/9163788

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

= 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

An indestructible bullet 2.00cm long is fired straight through a board that is 10.0cm thick. The bullet strikes the board with a speed of 470 m/s and emerges with a speed of 270 m/s. (To simplify, assume that the bullet accelerates only while the front tip is in contact with the wood.) a). What is the average acceleration of the bullet through the board? ________m/s^2b). What is the total time that the bullet is in contact with the board? (Enter the total time for the bullet to completely emerge from the board.) _________s
c.) What thickness of board (calculated 0.1 cm) would it take to stop the bullet, assuming that the acceleration through all boards is the same? ________cm

Answers

Answer:

a) $a=-7.4* 10^(-5)\ m/s^2$

b) $t=0.27* 10^(-3)\ s$

c)s=14.92 cm

Explanation:

Given that

u= 470 m/s

v = 270 m/s

s= 10 cm

We know that

$v^2=u^2+2as$

$270^2=470^2+2* a* 0.1$

$a=-7.4* 10^(-5)\ m/s^2$

v= u + a t

$270=470-7.4* 10^(-5)* t$

$t=0.27* 10^(-3)\ s$

To stop the bullet it means that the final velocity will be zero.

$v^2=u^2+2as$

$0^2=470^2-2* 7.4* 10^(-5) * s$

s=14.92 cm

A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?

Answers

Answer:

a) x = 5.48 10⁻² m and b) 0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

A = √ (2Em / k)

A = √ (2 0.12 / 80)

A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

Em = K = ½ m v²

Final

Em = Ke = ½ k x²

½ m v² = ½ k x²

x = √(m/k) v

x = 2 √(0.50 /800.0)

x = 0.05 m

Answer:

a) The greatest extension of the spring is 0.055 m

b) The spring compress 0.05 m

Explanation:

Please look at the solution in the attached Word file

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A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?O A. 6,658 ftOB. 25,396 ftOC. 7,282 ftOD. 23,219 ft​

Answers

Related Questions

Answers

Answers

Electrical Power:

Answers

Deceleration (a) =?

a = –3.576 m/s²

Time (t) =?

t = 8.75 s

Distance (s) =?

s = 136.89 m

Answers

Answers

Answers

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft