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olga55 [171]
3 years ago
13

150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag

on, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to_________
Physics
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

the friction force on this box is closest to 45.9 N.

Explanation:

given data

Weight of the box W = 150 N

accelerating uniformly = 3.00 m/s²

coefficient of kinetic friction = 0.400

coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force  that is express as

friction force = mass × acceleration ..............1

here mass = weight ÷ g

mass = \frac{150}{9.8} = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

friction force =  45.9 N

So, the friction force on this box is closest to 45.9 N.

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A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
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Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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Why the effect of gravitational force is more in liquid than in solid?​
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  • Gravitational forces between molecules is completely negligible compared to the em forces.

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Answer:

The most likely items to be used are;

Ultrasound and X-rays

Explanation:

A routine visit to a dentist consists of two areas of activities, including;

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b) Oral prophylaxis, and dental cleaning

The dental examination may involve the use of X-rays, which allow the detection of cavities between the teeth

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Question 1 of 10
Marianna [84]

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

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