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zepelin [54]
3 years ago
7

A person standing at the top of mountain aconcagua would be approximately 4.3 mi high. the radius of earth is 3959 mi. what is t

he distance to the horizon from this point?

Physics
2 answers:
Fed [463]3 years ago
6 0

Explanation :

It is given that,

The radius of the earth is, r = 3959 mi

From the attached figure, we have to find the distance to the horizon from this point.

So, using Pythagoras theorem:

AP^2=OA^2-OP^2

x^2=(R+h)^2-R^2

x^2=(3959+4.3)^2-(3959)^2

x^2=34065.89\ mi

x=184.57\ mi

The distance to the horizon from this point is 184.57 mi.

Hence, this is the required solution.

PilotLPTM [1.2K]3 years ago
5 0
Besides ambitious, it would be 184.6 mi
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How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0
2 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
A passenger in a train accelerating smoothly away from a station observes that a child’s yo-yo hanging by its string from a lugg
olchik [2.2K]

Answer:

1.73 m/s²

3.0 cm

Explanation:

Draw a free body diagram of the yo-yo.  There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.

Sum of forces in the y direction:

∑F = ma

T cos 10° − mg = 0

T cos 10° = mg

T = mg / cos 10°

Sum of forces in the x direction:

∑F = ma

T sin 10° = ma

mg tan 10° = ma

g tan 10° = a

a = 1.73 m/s²

Draw a free body diagram of the sphere.  There are two forces: weight force mg pulling down, and air resistance D pushing up.  At terminal velocity, the acceleration is 0.

Sum of forces in the y direction:

∑F = ma

D − mg = 0

D = mg

½ ρₐ v² C A = ρᵢ V g

½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g

3 ρₐ v² C = 8 ρᵢ r g

r = 3 ρₐ v² C / (8 ρᵢ g)

r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))

r = 0.030 m

r = 3.0 cm

3 0
3 years ago
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with
Dvinal [7]

Answer:

8 seconds

Explanation:

Answer:

Explanation:

Going up

Time taken to reach maximum height= usin∅/g

=3 secs

Maximum height= H+[(usin∅)²/2g]

=80+[(60sin30)²/20]

=125 meters

Coming Down

Maximum height= ½gt²

125= ½(10)(t²)

t=5 secs

6 0
3 years ago
A way of showing how energy moves through an ecosystem _________
Nataly_w [17]

Explanation:

Plants, as a autotrophs have chlorophyll to capture light energy from sun to make starch and sugar. Then, consumers eat plants, and the sugar is transferred to higher trophic level in a form of organic food. Nevertheless, energy is lost by uneaten food, indigestible food, unabsorbed food, excretory waste (eg co2) and heat loss by respiration.

7 0
3 years ago
Read 2 more answers
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