Answer:
This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h
24.29m/s or 87.4km/h
Explanation:
Let's find the acceleration of the car:
let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s
a = (vf-vi)/Δt
a = (16.67m/s-0)/8.0
a = 2.08m/s^2
Now we can use this acceleration to find vf in the second part:
50km/h is 13.89m/s
a = (vf-vi)Δt
vf = aΔt + vi
vf = 2.08m/s^2*5.0+13.89m/s
vf = 24.29m/s (87.4km/h)
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
Answer:
15 cm
Explanation:
= Diameter of the coin = 15 mm
= Diameter of the image of coin = 5 mm
= distance of the coin from mirror = 15 cm
= distance of the image of coin from mirror = ?
Using the equation
= - 5 cm
= radius of curvature
Using the mirror equation
= - 15 cm