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deff fn [24]
2 years ago
7

Select the star life cycle that is accurate.

Physics
1 answer:
Veseljchak [2.6K]2 years ago
8 0

Answer:

B according to the picture

Explanation:

Star of 3 stellar masses => red supergiant=>supernova=>neutron star

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The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
3 years ago
What to do if vehicle struck by lightning
agasfer [191]
Lighting flows around the outside of a truck, and the majority of the current flows from the cars metal cage into the ground below. It's not very safe to be in a car or truck during bad weather.
4 0
3 years ago
Which of the following is an example of acceleration?
Naddika [18.5K]

Answer:

c

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.

7 0
2 years ago
To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where
BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

\int E.dA = \frac{q}{\epcilon_0}

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

q = \int \rho dV

q = \rho * \pi r^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r}{2 \epcilon_0}

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

q = \int \rho dV

q = \rho * \pi r_0^2 *L

now using the equation of Gauss law we will have

\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}

now we will have

E = \frac{\rho r_0^2}{2 \epcilon_0 r}


7 0
3 years ago
3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them
zavuch27 [327]

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

F_k=\mu_kN

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

⇒ a = 0.1962 m/s^2

4 0
2 years ago
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