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dalvyx [7]
4 years ago
11

A hockey pack is set in motion across a frozen pond. If the ice friction and air resistance are neglected, the force required to

keep the pack sliding at constant velocity is A) Zero newtons. B) Equal to the weight of the pack. C) The weight of the pack divided by the mass of the pack. D) The mass of the pack multiplied by 9.8 meters per second per second. E) None of these.
Physics
1 answer:
Reil [10]4 years ago
3 0

Answer:

A) Zero newtos

Explanation:

To solve this exercise, let's use Newton's first and second law, in the first law it is established that every body is at rest or moving with constant speed if the external forces are zero. In the second law it is determined that the sum of the force is proportional to the acceleration of the body.

Let us apply our case, the body is at constant speed which implies that the sum of all external forces is zero. In addition, there is no friction, so we do not need a force that counteracts them.

Consequently, the force applied must be zero.  

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olga nikolaevna [1]
<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

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minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

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The minor measurement of the vernier scale is 0.02mm.

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2 years ago
Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
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consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

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A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.
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The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

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4vir4ik [10]

Answer:

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Explanation:

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I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

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