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2 years ago

3

Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the end

ing point.
Which describes where Manny most likely walked?
A. From the bottom of a hill to the top
B. On a circular nature trail
C. On a sidewalk from his house to the mail
D. From the beginning of a straight track to the end

2 answers:

Semenov [28]2 years ago

8 0

The correct answer is B on a circular nature trail.

We can tell this because the reference point at the start and end was the same because in a circular path the starting and ending are always at the same point.

In other case such as from bottom of the hill to the top the reference point can not be the same.

Similarly the reference point can not be the same for the starting and ending on the side walk or the straight track.

lesya [120]2 years ago

6 0

Answer:

it is B

Explanation:

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Answer:

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Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

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$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

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now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

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2 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

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Then, x_all = 150cm + 140cm = 290cm

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kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

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Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

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Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

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Explanation:

From the question we are told that

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So

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