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3 years ago

At which position would a person on Earth be able to view a solar eclipse? A B C D

Physics

1 answer:

sweet-ann [11.9K]3 years ago

5 0

The answer would be A, since at that position the Earth is fully/partially being engulfed in a shadow cast by the moon from the sun’s rays.

Hope this helps!

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(2)

katrin [286]

What if I don’t give you the answer

6 0

2 years ago

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)

v₀ₓ = 63.5 m/s

y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)

v₀y = 54.2 m/s

7 0

3 years ago

QUIZZ 12 + 14 + 66 - 87

svet-max [94.6K]

12 + 14 + 66 - 87

26+66-87

92-87

6 0

3 years ago

Read 2 more answers

A 2 kilogram mass is lifted 4 meters above the ground what is the change it gravitational energy

lianna [129]

Answer:

78.4 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location in the gravitational field.

The change in gravitational potential energy of an object is given by:

$\Delta U=mg\Delta h$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration due to gravity

$\Delta h$ is the change in height of the object

For the mass in this problem, we have:

m = 2 kg is the mass

$\Delta h = 4 m$ is the change in height

So, its change in gravitational energy is:

$\Delta U=(2)(9.8)(4)=78.4 J$

6 0

3 years ago

A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

NemiM [27]

Answer:

a) $F_H=776.952\ N$

b) $F_g=706.32\ N$

c) $v=5.4249\ m.s^{-1}$

d) $KE=1059.48\ J$

Explanation:

Given:

mass of the astronaut, $m=72\ kg$

vertical displacement of the astronaut, $h=15\ m$

acceleration of the astronaut while the lift, $a=\frac{g}{10} =0.981\ m.s^{-2}$

Now the force of lift by the helicopter:

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

$F_H-F_g=m.a$

where:

$F_H=$ force by the helicopter

$F_g=$ force of gravity

$F_H=72\times 0.981+72\times9.81$

$F_H=776.952\ N$

The gravitational force on the astronaut:

$F_g=m.g$

$F_g=72\times 9.81$

$F_g=706.32\ N$

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, $u=0\ m.s^{-1}$

using equation of motion:

$v^2=u^2+2a.h$

$v^2=0^2+2\times 0.981\times 15$

$v=5.4249\ m.s^{-1}$

Hence the kinetic energy:

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 72\times 5.4249^2$

$KE=1059.48\ J$

8 0

3 years ago

Read 2 more answers