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Citrus2011 [14]

2 years ago

10

What kind of star gives rise to a type i supernova?

1 answer:

larisa86 [58]2 years ago

3 0

a. It occures in binary systems where one of them is a whitedwarf

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Why are newtons second and third law important to everyday life?

Rzqust [24]

Because it always happens everyday, everytime, everyseconds...
Like typing on a keyboard, pushing a table. And if you're an engineer, you must able to measure it...
Bad english...

6 0

2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

2 years ago

a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit

Darina [25.2K]

Here is my step-by-step-work. Let me know if you have any questions! :)

3 0

2 years ago

An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

zhenek [66]

Answer:

2 m/s and -2 m/s

Explanation:

The object travels with an angle of

60.0°

with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).

If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:

$v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s$

If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:

$v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s$

5 0

2 years ago

Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m

Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

$V_{escape}=\sqrt{\frac{2GM}{R}}$

$V_{escape}=$ Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

$V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}$

$=15.88 \times 10^4$ m/s

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

3 0

2 years ago