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3 years ago

11

A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have falle

n after 3 seconds?

1 answer:

adell [148]3 years ago

5 0

The relevant equation to use here is:

y = v0 t + 0.5 g t^2

where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

y = 0 + 0.5 * 9.8 * 3^2

<span>y = 44.1 meters</span>

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Pushing, pulling is the answer

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3 years ago

The smallest detail visible with ground-based solar telescopes is about 1 arc second. How large a region (in km) does this repre

mestny [16]

Answer:

x = 727.5 km

Explanation:

With the conditions given using trigonometry, we can find the tangent

tan θ = CO / CA

With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun

D =150 10⁶ km (1000m / 1 km)

D = 150 10⁹ m.

We must take the given angle to radians.

1º = 3600 arc s

π rad = 180º

θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =

θ = 4.85 10⁻⁶ rad

That angle is extremely small, so we can approximate the tangent to the angle

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x = θ D

x = 4.85 10-6 150 109

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x = 727.5 km

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3 years ago

Plzzz answer this question correctly

tekilochka [14]

Answer:

changing the direction in which a force is exerted

8 0

3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

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$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0

3 years ago