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algol [13]
3 years ago
8

How much time will it take for a bug to travel 2 meters across the floor if it is traveling at 0.5 m/s?

Physics
1 answer:
7nadin3 [17]3 years ago
6 0
4 seconds to travel 2 meters :)
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6 0
4 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
A truck speed from rest to 120km/h in 13s. Find its acceleration. (convert to m/s)
ioda

answer:

2.5 m/s²

Explanation:

120 km/h = 120 ÷ 3.6 = 100/3 ≈ 33 m/s

a = (v2 - v1)/∆t = (33m/s - 0)/ 13s = 33/13 m/s²≈ 2.5 m/s²

4 0
3 years ago
Why is acceleration called vector quantity
MrRissso [65]
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7 0
3 years ago
a proton travelling along is x-axis is lowed by a niform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/
anzhelika [568]

Answer:

Magnitude of electric field is 1.06 x 10^5 V/m along negative X-direction

Explanation:

Given: initial velocity of proton = u = 3.5 x 10^6 m/s

final velocity of proton = v = 0 m/s

initial point l_i = 0.2 m and final point is l_f = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)

where q and m is the charge and mass of proton E is the electric field , l_i and l_f is the initial and final position of proton

on substituting the respected values we get,

1.023 x 10^-^1^4 = 9.6 x 10^-^2^0 x E

⇒ E = 1.06 x 10^5 V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x 10^5 V/m along negative X-direction

5 0
3 years ago
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