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Citrus2011 [14]
2 years ago
5

When an object is dropped from the top of a tall cliff, your teacher tells you to use -10 m/s2 for the approximate acceleration

due to gravity. What sign should the displacement of the object have and why?
A) The displacement should be positive, because it only moves in one direction.
B) The displacement should be positive, because this choice would make calculations easier
C) The displacement should be negative, because the negative represent a downward direction.
D) The displacement should be positive, because you can never have less than zero displacement.

The truck is described as travelling at 30mph to the west. The quantity described here is a(n)
A) speed
B) velocity
C) acceleration
D) kinetic energy
Physics
2 answers:
N76 [4]2 years ago
6 0

1- C

2-B

Those are your answers hope this helps :)


Romashka-Z-Leto [24]2 years ago
3 0

1. Answer: Option C

Displacement is a vector quantity. It has both magnitude as well as direction. Hence, it can be positive,  zero or negative. In the given question, downward direction is taken as negative. This we have come to know from the downward acceleration -10m/s^2. The negative sign indicates that downward direction is considered negative.

Now, since the object is dropped from the top. The displacement should be negative, because the negative represent a downward direction.

2. Answer: velocity

The truck is travelling at 30 mph to the west. From the unit mph (miles per hour), we know its a magnitude of velocity. West indicates the direction. Hence, the given quantity is a vector -velocity.

It is not speed because speed only indicates the magnitude and not the direction. speed is a scalar quantity.

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Acceleration is defined as the rate of change for which of the following
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A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro
Vlada [557]

The strength of the electric field on the point charge at this distance will be 4000 V/m.

<h3>What is the strength of the electric field?</h3>

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

To learn more about the strength of the electric field refer to the link;

brainly.com/question/15170044

#SPJ1

7 0
1 year ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
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