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Eduardwww
14 days ago
15

the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce

ntimeter, how far does the minute hand travel during a 42 minute period
Physics
1 answer:
Ostrovityanka [942]14 days ago
4 0

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

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Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

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5 days ago
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Paano nakatulong ang estratehikong lokasyon ng Pilipinas sa paghubog ng ating kasayasayan?​
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Answer:

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6 0
3 days ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [942]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0
7 days ago
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Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
Maru [1056]

Answer:

the maximum static friction force of the wall acting on the book (Increasing)

the normal force of the wall acting on the book (Decreasing)

the weight of the book (Constant)

Explanation:

According to Newton's third law of motion:

"Every action has an equal and opposite reaction"

In the scenario provided, Albert is pressing the book against the wall and subsequently decreases the force applied against the wall.

Let's evaluate all forces influencing the book in this situation.

1. Weight of the book acting downwards (y-axis)

2. Friction from the book against the wall acting upwards (y-axis)

3. Albert’s force exerted on the book against the wall (x-axis)

4. Normal force of the wall reacting to Albert’s applied force (x-axis)

As Albert eases off his force, the new scenario reads:

1. The weight remains constant as represented by W = mg

Since neither mass nor gravitational acceleration has changed, the weight exerted on the book remains the same.

2. As Albert reduces his force, the wall’s normal reaction force decreases correspondingly, following Newton's third law of motion.

3. Friction operates in response to the force applied to it. With a box resting on the floor, no friction acts upon it until it is dragged, at which point friction begins to manifest and rise until it reaches its maximum. Therefore, when Albert diminishes his force, the weight's pull will influence the book and the maximum static friction will rise to resist the book’s downward movement.

It should be noted that the maximum static friction is working to prevent movement of the book. With Albert's force reduced, but the weight of the book unchanged, maximum static friction increases to prevent downward movement.

7 0
8 days ago
An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center
Maru [1056]

Answer:

2023857702.507 m

Explanation:

f=\frac{GMm}{r^{2} }

Using Newton's law of gravitation:

G = gravitational constant

m_shrew = 50 g

m_elephant = 5 × 10^3 kg

r_earth = Earth's radius, 6400 km or 6,400,000 m

m_earth = Earth's mass

Equate the gravitational forces:

G m_shrew m_earth / r_earth^2 = G m_elephant m_earth / r^2

Cancel common terms on both sides:

m_shrew / r_earth^2 = m_elephant / r^2

Rearranged to solve for r^2:

r^2 = (m_elephant × r_earth^2) / m_shrew

Substituting the values:

r^2 = 4.096 × 10^{13}

Taking square root gives:

r = 2,023,857,702.507 m

4 0
15 days ago
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