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2 years ago

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

r semicircles. A greyhound can run around these turns at a constant speed of 15 m/s. What is its acceleration in m/s2
and in units of g?

1 answer:

Margarita [4]2 years ago

5 0

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

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A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

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2 years ago

The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Marat540 [252]

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

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2 years ago

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Calculate the volume of a liquid with a density of 5.45 g/ml and a mass of 65g

katrin [286]

Density=mass/volume
5.45g/ml=65g/V
V=65g/5.42g/ml
V=11.92ml

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2 years ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour miles/hou

dolphi86 [110]

Answer:

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

Explanation:

Acceleration is the change in velocity per unit time

a = ∆v/t

Given;

∆v = 50.0miles/hour - 0

∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour

∆v = 22.352m/s

t = 2.22 s

So,

Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

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2 years ago

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Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

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2 years ago