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3 years ago

If you saw a waxing gibbous moon what phase would you expect one week later?

1 answer:

6 0

Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.

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Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum

scoray [572]

Answer:

F = 2.30 10⁴ N

Explanation:

The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.

F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²

In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m

Let's calculate

F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²

F = 2.30 10⁴ N

The bond strength must be equal to or greater than this value

3 0

3 years ago

When a cyclist moves downhill without pedalling, what type of energy does he gain?

Vesna [10]

Answer:

He is gaining kinetic energy and losing potential energy

Explanation:

3 0

2 years ago

Read 2 more answers

The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si

JulijaS [17]

Answer:

0.75

Explanation:

$refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} }$

= 1.5

$refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$1.5 = \ \frac{ \sin(i) }{ \sin(30) }$

1.5 × ½ = sin(i)

$\sin(i) = 0.75$

5 0

3 years ago

the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w

4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

where K.E represents the kinetic energy,

m represents the mass,

v represents the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

1.10 $\times$ 10^42 = 1/2 $\times$ 3.26 $\times$ 10^31 $\times$ $v^{2}$

$v^{2}$ = (1.10 $\times$ 10^42 $\times$ 2) / (3.26 $\times$ 10^31)

v = 0.956 m / s.

6 0

3 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago