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Vinil7 [7]
3 years ago
6

Where on the Earth would you have to be to see the North Celestial Pole exactly halfway between your horizon and zenith?

Physics
1 answer:
mario62 [17]3 years ago
4 0

Answer:

The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.

From any location on Earth you see only half of the celestial sphere, the half above the horizon.

If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.

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A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.
g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle=34^{\circ}

Range=240 m

We know that Range =\frac{u^2sin2\theta }{g}

240=\frac{u^2sin(68)}{9.81}

u=50.39 m/s

(b)maximum height of projectile is given by

H=\frac{u^2sin^2\theta }{2g}

H=\frac{50.39^2(sin34)^2}{2\times 9.81}

H=40.46 m

3 0
3 years ago
Read 2 more answers
If it takes you 10 seconds to increase from 0 km/h to 50 km/h, what is your acceleration?
ZanzabumX [31]
Acceleration is change in velocity over change in time. Your Δv is +13.9, since you increased speed by 50 km/h which is 13.9 m/s, and your Δt is 10s. 13.9/10 = 1.39 m/s^2, the standard units for acceleration. Make sense?
4 0
3 years ago
Read 2 more answers
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the
arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

s_1=ut_1+\frac{1}{2}a_1 t_1^2

where:

u = 0 is the initial velocity (the car starts from rest)

t_1 = 20 s is the time elapsed in the 1st part

a_1=1.62 m/s^2 is the acceleration of the car in the 1st part

s_1 is the displacement of the car in the 1st part

Solving for s_1,

s_1=0+\frac{1}{2}(1.62)(20)^2=324 m

We can also find the velocity of the car after these 20 seconds using the equation:

v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

s_2 = (\frac{v_1 + v_2}{2})t_2

where:

v_1=32.4 m/s is the initial velocity at the beginning of the 2nd phase

v_2=0 is the final velocity (the car comes to a stop)

t_2=5 s is the time elapsed in the 2nd phase

Substituting,

s_2=\frac{32.4+0}{2}(5)=81 m

So, the total displacement of the car is

s=s_1+s_2=324+81=405 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh
coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0
2 years ago
Read 2 more answers
If a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right, wh
mr Goodwill [35]

The net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N. Details about net force can be found below.

<h3>How to calculate net force?</h3>

The net force of a body can be calculated by multiplying the mass of the body by its acceleration as follows:

Force = mass × acceleration

According to this question, a box with a mass of 8.0 kg is sitting on a frictionless surface and experiences an acceleration of 4.0 m/s2 to the right.

Net force = 8kg × 4m/s²

Net force = 32N

Therefore, the net force acting on a box of mass 8.0kg that experiences an acceleration of 4.0m/s² is 32N.

Learn more about net force at: brainly.com/question/18031889

#SPJ1

4 0
2 years ago
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