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RideAnS [48]
3 years ago
7

An imaginary dense star z has 1/5 the radius of earth, but 1000 times the earths. how much would a mass weighing 1.0n on earth w

eigh on star z
Physics
1 answer:
andrew11 [14]3 years ago
8 0
A 1-newton mass on earth would be 1000 newtons on star Z.

Funny enough, stars like this exist~! They're called "Neutron Stars."
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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
One form of energy can be _____ another type of energy.
kotegsom [21]

We can answer this one very quickly.  From the <em>Law of Conservation of Energy</em>, we know that "Energy can't be created or destroyed.".

So that only leaves us one way to complete the sentence in this question:

"One form of energy can be <em>transformed into</em> another type of energy. "  <em>(B)</em>

4 0
3 years ago
A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s aft
jok3333 [9.3K]

Answer:

4.8 m/s  

Explanation:

When she catches the train,

  1. They will have travelled the same distance.and
  2. Their speeds will be equal

The formula for the distance covered by the train  is

d = ½at² = ½ × 0.40t² = 0.20t²

The passenger starts running at a constant speed 6 s later, so her formula is

d = v(t - 6.0)

The passenger and the train will have covered the same distance when she has caught it, so

(1) 0.20t² = v(t - 6.0)

The speed of the train is

v = at = 0.40t

The speed of the passenger is v.

(2) 0.40t = v

Substitute (2) into (1)

0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t

Subtract 0.20t² from each side

0.20t² - 2.4t = 0

Factor the quadratic

t(0.20t - 2.4) = 0

Apply the zero-product rule

t =0     0.20t - 2.4 = 0

                   0.20t = 2.4

(3)                      t = 12

We reject t = 0 s.

Substitute (3) into (2)

0.40 × 12 = v

            v = 4.8 m/s

The slowest constant speed at which she can run and catch the train is 4.8 m/s.

A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.

7 0
4 years ago
What is the relationship between the wavelength of the peak (maximum) of the Gaussian curve and the temperature in degrees Kelvi
Tomtit [17]

Answer:

\lambda_{max}=\frac{b}{T}

Explanation:

Wien's displacement law states that the radiation of the black body curve for different temperatures will give peak values at different wavelengths and this wavelength is related inversely to the temperature.

Formally the law of Wien displacement states that the black body's spectral radiation per unit of wavelength, will give peaks at the wavelength of  \lambda_{max} which is given by the mathematical expression.

\lambda_{max}=\frac{b}{T}

Here, b is proportionality constant  with value of 2.897771955...\times 10^{-3} mK

The wavelength of the peak of the Gaussian curve is inversely related to temperature in degree kelvin.

3 0
3 years ago
Someone help me with letter d. and e. I’m supposed to solve for y.
Bumek [7]

Answer:

t=\frac{a}{\sqrt{1-\frac{y^2}{c^2}}}\\\frac{y^2}{c^2}=1-\frac{a^2}{t^2}\\y^2=\frac{c^2t^2-c^2a^2}{t^2}\\y=\frac{c}{t}\sqrt{t^2-a^2}

t=\frac{a}{\sqrt{1-\frac{v^2}{y^2}}}\\\frac{v^2}{y^2}=1-\frac{a^2}{t^2}\\\\\frac{v^2}{y^2}=\frac{t^2-a^2}{t^2}\\y^2=\frac{v^2t^2}{t^2-a^2}\\y=\frac{vt}{\sqrt{t^2-a^2}}

7 0
3 years ago
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