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Licemer1 [7]
3 years ago
15

How much force is need to accelerate a skier that is 66kg at a rate of 2 m/s2?

Physics
1 answer:
lubasha [3.4K]3 years ago
8 0

Answer:

132 N

Explanation:

newton's second law states that force = mass * acceleration

F = ma

plug in 66 kg for mass and 2 m/s^2 for acceleration

F = 66 * 2

F = 132 N

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Question 2 (5 points)
d1i1m1o1n [39]

Answer:

Frequency of sound wave = 198.83 hertz (Approx.)

Explanation:

Given:

Velocity of sound wave in air = 340 m/s

Wavelength = 1.71 meter

Find:

Frequency of sound wave

Computation:

Frequency = Velocity / Wavelength

Frequency of sound wave = Velocity of sound wave in air / Wavelength

Frequency of sound wave = 340 / 1.71

Frequency of sound wave = 198.8304

Frequency of sound wave = 198.83 hertz (Approx.)

6 0
3 years ago
How much energy is required to take 500g of ice at -40 degrees to water at room temperature (22 degrees Celsius)? Show work pls
Minchanka [31]

Answer: Given that:

Mass (m) = 20 g = 0.02 Kg,

temperature (T₁) = -10°C = -10+273 = 263 K

temperature (T₂) = 10°C = 10+273 = 283 K

Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)

                                                   = 0.02 × 4.187 × (283-263)

                                                Q = 1.67 KJ

Heat transferred is 1.67 KJ

Explanation:

3 0
3 years ago
ou are pushing a 20-kg box along a horizontal floor. Friction acts on the box. When you apply a horizontal force of magnitude 48
ziro4ka [17]

Answer:

The magnitude of the frictional force is 48.02 N

Explanation:

Mass of box = 20 kg

Weight of the box (Normal reaction) = mass × acceleration due to gravity = 20 ×9.8 = 196 N

Horizontal force applied = 48 N

Coefficient of friction = horizontal force ÷ normal reaction = 48 ÷ 196 = 0.245

Frictional force = coefficient of friction × normal reaction = 0.245 × 196 N = 48.02 N

5 0
3 years ago
Read 2 more answers
Placement.
stealth61 [152]

Answer:

kinetic energy will be equal to 0

Explanation:

this is because at final position velocity of body will become zero.

kinetic energy eill be 8 times

7 0
3 years ago
two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation
ad-work [718]

The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

  • k is the spring constant
  • m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

3 0
1 year ago
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