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3 years ago

9

In which phase is the full illuminated face of the Moon visible on Earth? A) New moon Eliminate B) Full moon C) Waning gibbous D

) Waxing crescent

1 answer:

Nitella [24]3 years ago

4 0

Full moon

New cant be seen

Gibbous is 3/4

Crescent is 1/4

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Where do we hear loud sound, at antinode or node?

tester [92]

Answer:

node

Explanation:

on the graph node is higher than antinode

so it can get or hear loud sounds faster

4 0

3 years ago

A boy throws a ball into the air at 10.2 m/s. Assuming that only gravity acts on the ball, how high does it rise, in m?

ozzi

Answer: 5.31 meters

Explanation: Use conservation of energy. Initial energy equals final energy. Initially, there is only kinetic energy (because height = 0 initially). At the end, kinetic energy equals 0 because at max height, there is max potential energy and the ball stops moving for a split second.

mgh = .5mv^2
Masses cancel out
gh = .5v^2
(9.8)(h) = .5(10.2^2)
Solve for h. h = 5.31 meters

5 0

3 years ago

An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

BlackZzzverrR [31]

Answer:

a) $s = 0.603\,m$ , b) $s = 2.412\,m$

Explanation:

a) The system ice cube-spring is modelled by means of the Principle of Energy Conservation. Let assume that height at the bottom is zero:

$U_{g} = U_{k}$

$m\cdot g \cdot s\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x^{2}$

$s = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \sin \theta}$

$s = \frac{(25\,\frac{N}{m} )\cdot (0.1\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 0.603\,m$

b) The distance travelled by the ice cube is:

$s = \frac{(25\,\frac{N}{m} )\cdot (0.2\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 2.412\,m$

7 0

2 years ago

Read 2 more answers

Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic inte

faltersainse [42]

Answer:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A

Explanation:

Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.

5 0

3 years ago

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

Answer:

a) y= 3.5 10³ m, b) t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

y₁ = y₀ + v₀ t + ½ a t²

y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

y₁ = ½ 16 10²

y₁ = 800 m

At the end of this stage it has a speed

v₁ = vo + a₁ t₁

v₁ = 0 + 16 10

v₁ = 160 m / s

Stage 2

y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

y₂ = 800 + 150 5 + ½ 11 5²

y₂ = 1092.5 m

Speed is

v₂ = v₁ + a₂ t

v₂ = 160 + 11 5

v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

v₃² = v₂² - 2 g y₃

0 = v₂² - 2g y₃

y₃ = v₂² / 2g

y₃ = 215²/2 9.8

y₃ = 2358.4 m

The total height is

y = y₃ + y₂

y = 2358.4 + 1092.5

y = 3450.9 m

y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

v₃ = v₂ - g t₃

v₃ = 0

t₃ = v₂ / g

t₃ = 215 / 9.8

t₃ = 21.94 s

The time to climb is

$t_{s}$ = t₁ + t₂ + t₃

t_{s} = 10+ 5+ 21.94

t_{s} = 36.94 s

The time to descend from the maximum height is

y = v₀ t - ½ g t²

When it starts to slow down it's zero

y = - ½ g t_{b}²

t_{b} = √-2y / g

t_{b} = √(- 2 (-3450.9) /9.8)

t_{b} = 26.54 s

Flight time is the rise time plus the descent date

t = t_{s} + t_{b}

t = 36.94 + 26.54

t =63.84 s

t = 64 s

3 0

3 years ago