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3 years ago

8

A runner starts at the start line runs around a 400 m track two times ending up back at the store now what is the runners distan

ce and displacement

1 answer:

Marta_Voda [28]3 years ago

7 0

Answer:

Distance is 800m and displacement is 0m

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You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

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3 years ago

Read 2 more answers

1. At a location in Europe, it is necessary to supply 1000 kW of 60-Hz power. Only power sources available operate at 50 Hz. It

Klio2033 [76]

Answer:

Explanation:

From the given information:

The speed of a synchronous motor in relation to its frequency can be represented with the formula:

$n_{sm}= \dfrac{120f_{se}}{P}$

where,

the electrical frequency $f_{se}$ is measured in Hz

the number of poles = P

For us to estimate the number of poles to have 50 Hz - 60 Hz Power, then we need to relate the frequencies of the above equation.

i.e

$\dfrac{120(50 \ Hz)}{P_1}= \dfrac{120( 60 \Hz)}{P_2} \\ \\ \dfrac{6000 \ Hz}{P_1}= \dfrac{7200 \ Hz}{P_2} \\ \\ \dfrac{P_2}{P_1}=\dfrac{7200}{6000} \\ \\ \\ \dfrac{P_2}{P_1}= \dfrac{12}{10}$

Thus, we can conclude that 10 poles synchronous motor is attached with 12 poles synchronous generator in order to convert 50 Hz to 60 Hz power.

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3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago