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3 years ago

If in 30 mins Alex runs to a store that is 4 km away, what is her average speed in km/h

Physics

1 answer:

masha68 [24]3 years ago

6 0

15 km 30 divided by 4 is 7.5 km in 30 min times that by 2 the answer is 15

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A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

4 0

3 years ago

High energy waves have short wave lengths, true or false?

Anna71 [15]

True, the wavelength dies down due to high frequency and low amptitude.

5 0

3 years ago

A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m

konstantin123 [22]

Answer:

The torque on the loop is $2.4 \times 10^{-2}$ Nm

Explanation:

Given:

Current $I = 1.6$ A

Magnetic field $B = 0.30$ T

Area of loop $A = 0.14$ $m^{2}$

Angle between magnetic field and area vector $\theta =$ 21°

Form the formula of torque in case of magnetic field,

г $= MB \sin \theta$

Where $M =$ magnetic moment

$M = IA$

г $= IAB \sin 21$

г $= 1.6 \times 0.30 \times 0.14 \times 0.3583$

г $=2.4 \times 10^{-2}$ Nm

Therefore, the torque on the loop is $2.4 \times 10^{-2}$ Nm

7 0

4 years ago

All of the follow are ways of measuring volume execpt:

bogdanovich [222]

A). Scale.
You CAN measure volume with all the following except a scale.

6 0

4 years ago

One of the fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974, was clocked at 100.8 mi/hr. If a pitch

lawyer [7]

<u>Answer:</u>

The ball fall vertically 2.69 ft by the time it reached home plate 60.0 ft away.

<u>Explanation:</u>

Fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974 = 100.8 mi/hr = 44.8 m/s

The horizontal distance to home plate = 60.0 ft = 18.288 m

We have the horizontal velocity = 44.8 m/s

So time taken = 18.288/44.8 = 0.408 seconds.

The distance traveled by baseball vertically is found out by equation $s=ut+\frac{1}{2} at^2$

Here u =0m/s, a = 9.81 $m/s^2$ and t = 0.408 s

Substituting

$S = 0*0.408+\frac{1}{2} *9.81*0.408^2 = 0.82 m\\ \\S = 0.82 m= 2.69 ft$

So vertical distance traveled = 0.82 m = 2.69 ft

7 0

3 years ago

Read 2 more answers