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17 days ago

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

e placed
charges of +3.0 μC and -3.0 μC. What is the potential (relative to infinity) at the midpoint of the square? (k = 1/4πε0 = 9.0 × 109 N · m2/C2)

A) 1.0 × 104 V B) 1.0 × 105 V C) 1.0 × 106 V D) 0 V

E) infinite Answer: B

Physics

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A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav

kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0

3 months ago

Read 2 more answers

If you are anchored in a fixed spot, and a set of six waves pass underneath you during a 60 second time interval, what is the wa

Ostrovityanka [3204]

Answer:

Explanation:

Within a duration of 60 seconds, six waves are observed.

With a total of 6 waves,

this equates to 3 wavelengths.

As a result,

the period for each wavelength is calculated as 60 divided by 3.

Thus, period = 20 seconds.

According to the frequency-period relationship,

f = 1 / T

f = 1 / 20

f = 0.05 Hz

5 0

3 months ago

A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part

ValentinkaMS [3465]

answer:

$Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\
Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\ i.e\\
4f_{f}=weight =mg\\
f_{f}=\frac{55x9.8}{4}\\
=134.75N$

5 0

3 months ago

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

serg [3582]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

6 0

3 months ago