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3 years ago

11

Black holes are the final stage of what type of star?

2 answers:

creativ13 [48]3 years ago

7 0

Answer:

the neutron star becomes a black hole

kap26 [50]3 years ago

3 0

Answer:

neutron stars

Explanation:

not really sure of an explanation, but maybe try looking up for "What type of star ends up as a blackhole?" should help

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A magnetic field would be produced by a beam of

kumpel [21]

3 protons should be your answer

4 0

3 years ago

How does an objects size affect the gravity exerted by that object

egoroff_w [7]

Answer:

The size of an object is directly proportional to the gravity

Explanation:

The size of an object has significant impact on the gravity exerted by such a body.

The more massive a body is, the larger the gravity it exerts.

The reason for this is because of the newton's law of universal gravitation.

It states that "the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
As such, gravity is directly proportional to mass

4 0

3 years ago

A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever

kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier $(V_s)$ = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave = $(V_s - V_w)$

where:

$V_s$ = velocity of the skier

$V_w$ = velocity of the wave

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s-V_w)T$

$\lambda = (V_s-V_w) 0.450m$ ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = $(V_s + V_w)$

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s+V_w)T$

$\lambda = (V_s+V_w) 0.342m$ ------------------> Equation 2

Equating equation (1) and equation (2) and substituting $V_s$ = 14.3 m/s ; we have:

$(V_s-V_w) 0.450m = (V_s-V_w) 0.342m$

$0.450m(V_s)-0.450m(V_w) = 0.342m(V_s)+0.342m(V_w)$

Collecting the like terms; we have:

$0.450m(V_s) - 0.342m(V_s) = 0.342m(V_w)+0.450m(V_w)$

$(V_s)(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.108m = (V_w)0.792m$

$1.5444m^2/s = (V_w)0.792m$

$(V_w) = \frac{1.5444m^2/s}{ 0.792m}$

$(V_w) = 1.95 m/s$

b)

The Wavelength of the wave can be calculated using : $( \lambda }) = (V_s-V_w) 0.450m$

$({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)$

$(\lambda) = (12.35)0.450m$

$(\lambda)= 5.5575 m$

λ ≅ 5.56 m

5 0

3 years ago

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran

marshall27 [118]

Answer

a) For the rock

$\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}$

$2sin\thetacos\theta = \dfrac{sin^2\theta}{2}$

$2cos\theta = \dfrac{sin\theta}{2}$

$tan\theta = 4$

$\theta = tan^{-1} 4$

$\theta = 76^0$

b) $\theta = 45^0$ for maximum range

$\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}$

$\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}$

$\dfrac{d_{max}}{d}=2.129$

$d_{max}=2.129 d$

c) The value of θ is the same on every planet as g divides out.

5 0

3 years ago

If the speed of a wave is 50 m/s and the period of time between waves is 2 seconds. what is the frequency?

Rashid [163]

Answer:

25Hz

Explanation:

3 0

3 years ago