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kakasveta [241]
3 years ago
6

Faiyez wrote the problem below

Physics
1 answer:
Ilya [14]3 years ago
6 0

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

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Akio draws the ray diagram shown.
Inessa [10]

Answer: Move the small car so it appears on the left side of the lens.

Explanation:

Because the lens is reflective the small car would apear on the same side as the normal car.

Hope this helps :)

3 0
3 years ago
if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a
Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

4 0
2 years ago
The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t
yuradex [85]

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

6 0
3 years ago
Oo8. The diagram below models Mercury and Venus orbiting the sun.
katrin [286]
I’m so sorry I been trying to figure this out for sun
3 0
2 years ago
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How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m
Mars2501 [29]

Answer:

F=2496 N

Explanation:

Given that,

Mass of SUV, m = 1600 kg

Initial speed, u = 0

Final speed, v = 25 m/s

Distance, d = 200 m

We need to find the net force. Firstly, let's find acceleration using equation of motion.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(0)^2}{2\times 200}\\\\a=1.56\ m/s^2

Net force, F = ma

F=1600\times 1.56\\\\F=2496\ N

So, the net force is 2496 N.

6 0
3 years ago
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