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3 years ago

A speedboat is moving at a rate of 45 km per hour travels a distance of 27 km. How long did it take to go to 27 km.

Physics

1 answer:

8090 [49]3 years ago

8 0

Answer:

I think it is 1.67

Explanation:

I multiply the 45 and 27 cause it says rate of which means to multiply.

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Transverse waves travel with a speed of 20.0 m/s on a string under a tension of 6.00 N. What tension is required for a wave spee

marta [7]

Based on the sped of the waves and the tension as well as the needed wave speed, the required tension is 13.5 N.

<h3>What is the required tension?</h3>

Given the initial tension and speed, the tension that is required can be found by the formula:

= Initial tension x (Required speed / Initial speed)²

Solving gives:

= 6 x (30 / 20)²

= 6 x 9/4

= 13.5 N

In conclusion, the tension required is 13.5N.

Find out more on the tension on a wire at brainly.com/question/14290894.

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3 0

2 years ago

A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of

Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

$K_1 = K_2 + P + W$

then, the work done by the engine is:

$W = K_1 - K_2 - P$

The formulas for the Kinetic and potential energy are:

$K=\frac{1}{2}mV^2\\P=mgh$

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

$W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh$

Replacing the values:

$W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J$

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

$t=\frac{3200m}{20m/s}=160s$

the power delivered by the car's engine was:

$power=\frac{work}{time}=\frac{1690400J}{160s}=10565W$

8 0

4 years ago

in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube

tekilochka [14]

Answer:

The viscosity is 1.30 x 10^-3 deca poise.

Explanation:

Volume per minute, V = 10^-5 m^3

Volume per second, V = 1.67 x 10^-7 m^3

density, d = 800 kg/m^3

radius, r = 0.05 cm

Length, L = 0.5 m

Height, h = 60 cm

Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa

Use the formula of rate of flow

$V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 = 3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise$

6 0

3 years ago

An electron with a velocity of 10^7 m/s enter into a region of magnetic flux density of 0.10T,the angle between the direction of

wolverine [178]

Answer:

F = q v B sin Θ describes force on the electron

F = m v^2 / R describes force required to keep electron in circular path

q v B sin Θ = m v^2 / R

R = m v / (q B sin Θ)

R = 9.1E-31 * 1E7 / (.1 * 1.6E-19 * .423)

R = 9.1 / 6.77 * E-3 = .00134 m = .134 cm

6 0

2 years ago

Calculate the energy transferred by an appliance using mains electricity (230V) if the charge is 150C. Give your answer in kiloj

Otrada [13]

The energy transferred by the appliance using mains electricity is 17.3 KJ

<h3>Data obtained from the question </h3>

Potential difference (V) = 230V
Charge (Q) = 150 C

Energy (E) =?

<h3>How to determine the energy transferred </h3>

The energy transferred can be obtained as follow:

E = ½QV

E = ½ × 150 × 230

E = 75 × 230

E = 17250 J

Divide by 1000 to express in kilojoules

E = 17250 / 1000

E = 17.3 KJ

Learn more about energy stored in a capacitor:

brainly.com/question/14739936

8 0

2 years ago