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3 years ago

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

he moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 14 m/s at an angle 20 ∘ above the horizontal. For how much more time was the ball in flight?

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

$T=\frac{2uSin\theta }{g}$

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

Tm - Te = 5.84 - 0.98 = 4.86 s

So, it takes 4.86 s more time of flight on moon than the earth.

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Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

Black_prince [1.1K]

The initial momentum of the system can be expressed as,

$p_i=m_1u_1+m_{2_{}}u_2$

The final momentum of the system can be given as,

$p_f=m_1v_1+m_{2_{}}v_2$

According to conservation of momentum,

$p_i=p_f$

Plug in the known expressions,

$\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}$

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

$\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}$

Thus, the final velocity of second mass is 2.99 m/s.

3 0

1 year ago

A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

const2013 [10]

Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

$\theta=12.19^{\circ}$

Therefore the angle will be 12.19° .

4 0

3 years ago

To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi

PolarNik [594]

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

$Z = \dfrac{X - \mu}{\sigma})$

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

Learn more about z-score here:

brainly.com/question/21262765

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2 years ago

The speed of light in a solid is 1.24 x 108 m/s. <br> Calculate the index of refraction

Dahasolnce [82]

Answer:

125

Explanation:

7 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago