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stira [4]
2 years ago
12

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

Physics
2 answers:
xxMikexx [17]2 years ago
8 0

Let's use Newton's 2nd law of motion:

                                     Force = (mass) x (acceleration)

                                     Force = (68 kg) x (1.2 m/s²) =  81.6 newtons .

labwork [276]2 years ago
5 0
f=m*a

68*1.2=81.6

f=81.6
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How is the volume flow rate of water out of the tank, dVdt, related to the flow speed v ? Express your answer in terms of some,
Rainbow [258]

Answer:

\frac{dV}{dt}= \frac{\pi d^2}{4}v

Explanation:

The rate of volume flow out of tank can be expressed as:

\frac{dV}{dt} = A\frac{dL}{dt}

where,

dV/dt = Volume flow rate

A = Cross-sectional area of outlet = πd²/4

d = diameter of circular outlet

dL = Displacement covered by water

dt = time taken

but we know that:

Velocity = υ = displacement/time = dL/dt

Substituting the values of "dL/dt" and "A" in the equation, we get:

\frac{dV}{dt} = \frac{\pi d^2}{4}v

This is the expression for volume flow rate dV/dt, on terms pf v, d.

8 0
3 years ago
What is a lot like gases but the atoms are different because they are made up of free electrons and ions of elements
34kurt

plasmas  are a lot like gases



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7 0
2 years ago
On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se
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It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.
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3 years ago
Read 2 more answers
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
2 years ago
Five men push a stalled car with an average force of 400 N per person. Find the mass of the
timama [110]

The mass of the car is 2000 kg

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

In this problem, we have:

a=1 m/s^2 is the acceleration of the car

Each person applies a force of 400 N, and there are five men, so the total force applied is

\sum F = 5\cdot 400 N = 2000 N

Therefore, the mass of the car is:

m=\frac{\sum F}{a}=\frac{2000 N}{1 m/s^2}=2000 kg

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

4 0
3 years ago
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