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3 years ago

9

A golfer takes two putts to get his ball into the hole he is on green. The first putt displaces the ball 4.5m east, and the seco

nd 8.27 m south. What displacement would have been needed to get the ball into the hole on the first putt.

1 answer:

LenaWriter [7]3 years ago

8 0

I think its the one going it at 4.5m

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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d

QveST [7]

Answer:

The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

(target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

(10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

(target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

-94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0

3 years ago

A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

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3 years ago

The accepted value for the acceleration due to gravity is 9.8 m/s2. What factors do you think are responsible for the difference

kakasveta [241]

It’s hard to perfectly measure the distance something travels, as well as the exact time it takes, making the results have some variation.

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3 years ago

Why do clouds tend to form over land with a sea breeze and over water with a land breeze?

Masja [62]

Because the waves in the water with the fan like system.

6 0

3 years ago

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106

LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = \frac{E_1q_1}{q_2}$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0

3 years ago