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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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A small ball is attached to one end of a massless, rigid rod. The ball and the rod revolve in a horizontal circle with the other

nikklg [1K]

Explanation:

<em>i </em><em>don't </em><em>have </em><em>keyboard </em><em>of </em><em>mathametics </em><em>so </em><em>i </em><em>solve </em><em>this</em>

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3 years ago

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Over [174]

The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:

$v_f^2-v_i^2 = 2ax$

Where,

$v_{f,i} =$ Final and initial velocity

a = Acceleration

x = Displacement

For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore

$v_f^2 = 2hg$

$v_f = \sqrt{2hg}$

Our values are given as

$h = 70km = 70*10^3m$

$g = 2m/s^2$

Replacing we have that,

$v_f = \sqrt{2hg}$

$v_f = \sqrt{2(70*10^3)(2)}$

$v_f = 529.15m/s$

Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s

6 0

3 years ago

What is carnot engine.express the relation for its efficiency?

patriot [66]

Answer:

Explanation:

The Carnot cycle is a special case of a thermodynamic cycle that produces an ideal gas and consists of two isothermal processes and two adiabatic processes. This cycle is a theoretical solution given by Sadi Karnot to refine heat engines for their efficient use.

The formula for the coefficient of efficiency is:

η = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁

Where Q₁ is is the amount of heat of the heater supplied to the working body and Q₂ is the amount of heat that the working body transfers to the refrigerator according to this T₁ is the temperature of the heater T₂ is the temperature of the refrigerator.

This formula provides a theoretical limit for the maximum value of the coefficient of efficiency of heat engines.

God is with you!!!

6 0

4 years ago

Read 2 more answers

Question 1 (1 point)

MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work $W$ done by the frictional force is

$W= Fd$ .

Now, $F = 60N$ and $d =10m$ ; therefore,

$W= (60N)(10m)$

$\boxed{W = 600J.}$

Hence, the work done by friction is 660J.

7 0

3 years ago

A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met

AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

$K.E_i +P.E_i= K.E_f + P.E_f$

$\dfrac{1}{2}kx^2 = mgh$

$\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h$

$\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h$

$156.8 = 44 \times 9.8 \times h$

$h = \dfrac{156.8}{44 \times 9.8}$

h = 0.364 m

h = 36.4 cm

7 0

3 years ago