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hjlf
3 years ago
15

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

Physics
2 answers:
DiKsa [7]3 years ago
4 0
<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
Alex73 [517]3 years ago
3 0

Answer:

The force needed to accelerate a 68 kilogram-skier at a rate of 1.2ms2 is 81.6 Net forces

Explanation:

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According to cell theory which of the following is not true?
Anettt [7]

Answer:

first one

Explanation:

4 0
3 years ago
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A mover loads a crate onto a truck bed 1.6m from the street using a ramp that is 4.6m long. What is a mechanical advantage?
Ne4ueva [31]

Answer:

Mechanical advantage = 2.875

Explanation:

Given:

A diagram is shown below for the above scenario.

Length of ramp (Effort arm) = 4.6 m

Height of truck bed ( Resistance length) = 1.6 m

Mechanical advantage (MA) is the ratio of effort arm and resistance length.

So, mechanical advantage is given as,

MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875

6 0
3 years ago
The acceleration due to gravity on the Moon's surface is
Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

g'=\frac{g}{6}

We know that weight of an object on Earth is,

w = m\times g

m = \frac{w}{g}

Similarly, weight on Moon will be

w' = m\times g'

w' = \frac{w}{g}\times\frac{g}{6}

w' = \frac{300}{6}

w' = 50

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0
3 years ago
Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a1x and a2x. find t
Serga [27]

we know that center of mass is given as

r = (m₁ r_{1x} + m₂ r_{2x})/(m₁ + m₂)

taking derivative both side relative to "t"

dr/dt = (m₁ dr_{1x}/dt + m₂ dr_{2x}/dt)/(m₁ + m₂)

v = (m₁ v_{1x} + m₂ v_{2x})/(m₁ + m₂)

taking derivative again relative to "t" both side

dv/dt = (m₁ dv_{1x}/dt + m₂ dv_{2x}/dt)/(m₁ + m₂)

a= (m₁ a_{1x} + m₂ a_{2x})/(m₁ + m₂)

3 0
3 years ago
Read 2 more answers
Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

Δx = v₁t₁

Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

4 0
3 years ago
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