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igor_vitrenko [27]
3 years ago
9

Using pressure and density of fluids.

Physics
1 answer:
mario62 [17]3 years ago
7 0

snow shoes.

shoes with large areas in contact with the snoow. reduces the pressure on the snow, and alows you to sink less. or to walk on the surface.


Stilettp high heels would have the opposite effect

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Laura and Elana are discussing how to solve the following problem: "A canary sits 10 m from the end of a 30-m-long clothesline,
horrorfan [7]

Answer:

f=2.236\ Hz

Explanation:

Given:

Length of a rope,l=30\ m

Position of Canary on the rope from one end, l_c=10\ m

Position of Grackle on the rope from another end, l_g=5\ m

Tension in the rope, F_T=200\ N

linear mass distribution on the rope, \mu=0.1\ kg.m^{-1}

We have for the speed of wave on the string:

v^2=\frac{F_T}{\mu}

v^2=\frac{200}{\0.1}

v=44.7\ m.s^{-1}

<em>For canary to be undisturbed we need a node at this location.</em>

<em>Also, at the end close to Canary there must be a node to avoid any change in pattern of vibration.</em>

So,

the distance between Canary and the closer end must be equal to half the wavelength.

\frac{\lambda}{2} =10\ m

\Rightarrow \lambda=20\ m

∴Wavelength of wave to be produced = 20 m. This will give us nodes at the multiples of 10 and anti-nodes at the multiples of 5.

Now, frequency:

f=\frac{v}{\lambda}

f=\frac{44.7}{20}

f=2.236\ Hz

3 0
3 years ago
Urgent! please help!
kap26 [50]

_{94}^{239}Pu+ _{0}^{1}n--->_{40}^{100}Zr+_{x}^{y}Element+2_{0}^{1}n\\ \\ 94+0=40+x+0, x= 54,\\ \\ 239+1 = 100+y+2*1, y=138\\ \\_{54}^{138}Element= _{54}^{138}Xe

Answer is A.

8 0
3 years ago
Then it is called
DiKsa [7]

Answer:

It is called force of friction

Explanation:

The force of friction is a force that acts between two objects whose surfaces are in contact with each other.

Consider the typical case of an object sliding along a certain surface. There are two types of frictions:

- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:

F_f = F

However, this force of static friction has a maximum value, which is given by

F_{max} = \mu_s N

where

\mu_s is the coefficient of static friction

N is the normal reaction exerted by the surface on the object

So, when F becomes greater than F_{max}, the static friction is no longer able to balance the push applied, and the object will start sliding forward.

- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by

F_f = \mu_k N

where

\mu_k is the coefficient of kinetic friction, and its value is generally smaller than \mu_s. The direction of this force is also opposite to the direction of motion of the object.

8 0
2 years ago
An is holding an ice cube. What causes the ice to melt?
ivann1987 [24]

Answer: A... Thermal energy from the ice is transferred to the air.

Explanation: because I just know.

6 0
2 years ago
Science Seminar Question: Why did Vehicle 2 fall off the cliff in Claire's test of the collision scene but Vehicle 2 did not fal
asambeis [7]

Complete Question:

Check the file attached to get the complete question

Answer:

In the film Ice word Revenge, vehicle 2 did not fall of the cliff because, Weight_{vehicle 1} < Weight_{vehicle 2} but in Claire's test, vehicle 2 off the cliff because Weight_{vehicle 1} \geq Weight_{vehicle 2}

Explanation:

In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff.  In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.

8 0
3 years ago
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