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3 years ago

What will one see during an annular eclipse?

1 answer:

7 0

Partial eclipse, Annular eclipse, Total Eclipse and Hybrid Eclipse are the four different types of the eclipses. When the Sun and Moon are exactly in line with the Earth, the annular eclipse occurs. The new moon is invisible from the earth and it is silhouetted against the sun, this can only be seen in annular eclipse. Annular word means ring shaped, we can see a dark ring of fire in annular eclipse. It has five different stages that are first contact, second contact, maximum eclipse, third contact and 4th contact.

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A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are

Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, $\theta=30^{\circ}$

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)$

$v = 0$

$Fd\ cos\theta=\dfrac{1}{2}m(u^2) F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N$

The force required to push to stop the car is 288.67 N

3 0

3 years ago

How far will a 70N crate is move with 3500J of work

r-ruslan [8.4K]

Work = Force * distance
Force = 70 N
Work = 3500 J

3500 = 70d
d = 3500/70 = 50 m

6 0

3 years ago

Read 2 more answers

The relationship between a cathode and an anode involves

astra-53 [7]

It involves electrons.

The cathode is the electrode where electron deficient ions move to.

While the anode is electrode where electron excess ions move to.

So the relationship between Cathode and Anode involves electrons.

C.

5 0

3 years ago

[Assume g = 10m/s?] If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her acceleration. (Ans: 0)

DerKrebs [107]

Answer:

Her acceleration is 0 m/s²

Explanation:

We note that the motion of the girl is on a straight road, therefore;

The vertical acceleration (e.g. due to gravity, <em>g)</em> on the horizontal motion = 0

The horizontal acceleration, a = (Change in velocity, Δv)/(Change in time, Δt)

For uniform velocity, the change in velocity, Δv = 0

Therefore, fore any change in time, Δt, we have;

a = Δv/Δt

Her acceleration, a = 0/Δt = 0

Her acceleration, a = 0 m/s²

7 0

3 years ago

Calculate the most charge that a 100pF capacitor (with a 1.0mm plate separation) can store if the capacitor breaks down with an

Misha Larkins [42]

Answer:

0.6 μC

Explanation:

C = capacitance of the capacitor = 100 x 10⁻¹² F

d = separation between the plates of capacitor = 1 mm = 1 x 10⁻³ m

E = Electric field = 6 x 10⁶ N/C

Q = Amount of charge

V = Potential difference

Potential difference is given as

V = E d

Amount of charge stored is given as

Q = CV

hence

Q = C E d

inserting the values

Q = (100 x 10⁻¹²) (6 x 10⁶) (1 x 10⁻³)

Q = 6 x 10⁻⁷ C

Q = 0.6 μC

5 0

3 years ago