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3 years ago

A snowball starting at rest rolls down a hill and reaches 5 m/s. If the hill is

Physics

1 answer:

Lostsunrise [7]3 years ago

4 0

Answer:

The acceleration of the snowball is 0.3125

Explanation:

The initial speed of the snowball up the hill, u = 0

The speed the snowball reaches, v = 5 m/s

The length of the hill, s = 40 m

The equation of motion of the snowball given the above parameters is therefore;

v² = u² + 2·a·s

Where;

a = The acceleration of the snowball

Plugging in the values, we have;

5² = 0² + 2 × a × 40

∴ 2 × 40 × a = 5² = 25

80 × a = 25

a = 25/80 = 5/16

a = The acceleration of the snowball = 5/16 m/s².

The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .

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2 years ago

6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air

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From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

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2 years ago

While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba

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3 years ago

Read 2 more answers

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp

In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

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3 years ago

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