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denis23 [38]
3 years ago
10

2 % of freshwater is locked in ______.

Physics
1 answer:
oksano4ka [1.4K]3 years ago
5 0
According to the internet over 2% of earths fresh water is unavailable: locked up in glaciers, polar ice caps, atmosphere, and soil.
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A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic
ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

I_i = Initial angular momentum = 275\ kgm^2

\omega_i = Initial angular velocity = 14 rev/min

I_f = Final angular momentum

\omega_f = Final angular velocity

Here the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s

The final angular velocity is 0.2932 rad/s

7 0
3 years ago
What is the current through a 11v bulb with a power of 99w
Murrr4er [49]

Answer:

9.01amp

Explanation:

Power = V^2/R

Given that v = 11volts, P = 99watts

99 = 11^2/R

11×11 = 99R

121= 99R

R = 121/99

R= 1.22ohms

From ohms Law; V = IR

11volts = I × 1.22ohms

I = 11/1.23

I = 9.01 amp

8 0
3 years ago
A circular loop of wire lies flat on a level table top in a region where the magnetic field vector points straight upward. The m
jarptica [38.1K]
The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.
4 0
3 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
During science class, Ava and Justin were discussing the rotation of the Sun and debating over which model would best illustrate
agasfer [191]

Answer:

it would b B

Explanation:

3 0
3 years ago
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