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AleksandrR [38]
3 years ago
6

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The side

s of the trunks that face each other are separated by 1.1 m. A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is 0.9 m above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is 1.7 m above the ground. Finally, he leaps to the other tree, now landing at a spot that is 2.3 m above the ground. What is the magnitude of the squirrel's displacement?
Physics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

2.55 m.

Explanation:

Displacement is the shortest distance between initial and final position . In the present case this distance forms the hypotenuse of the triangle having base equal to 1.1 m and perpendicular equal to 2.3 m .

Hypotenuse H then

H² = 1.1² + 2.3²

H = 2.55 m.

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4 years ago
Can someone help with these
dedylja [7]

6.  0N.  This questions requires understanding of how friction functions.  Friction is a resistive force, meaning it opposes the direction of any applied or unbalanced forces.  The box in the question experiences no horizontal force, so there is no resistive force in response to it, making it 0N.


7.  This question tests your understanding of static friction.  Static friction only applies when an applied or unbalanced force is applied to an object which does not move.  The static friction always equals the magnitude of the applied or unbalanced force component parallel to the surface which the object rests on.  The question states that the crate starts to move only when the applied force exceeds 313N, so we use this value to determine the force of static friction.  The additional info in the question pertaining to when the crate is moving is irrelevant when determining static friction (only relevant if determining kinetic friction).  Knowing this we solve for the weight of the crate:

F = mg

F = (45)(9.8)

F = 441N = Normal Force

The weight of the crate is also equal to the Normal Force since the object rests on a horizontal surface and the applied force is horizontal as well.  In this question, since the object is not moving at 313N of applied force, the magnitude of static friction equals the applied force:

Ff = μs * Fn

(313) = μs (441)

0.71 (rounded) = μs


4 0
3 years ago
Read 2 more answers
Why is the absolute magnitude of some stars greater than their apparent magnitude
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<span>answer under the link: http: //briskrange.com/7gAl</span>
6 0
4 years ago
What are the three rules for writing the electron configuration of elements
astraxan [27]

<em></em>

Answer:

<u><em>The aufbau principle</em></u>

<u />

<u><em>The Pauli exclusion principle</em></u>

<u><em></em></u>

<u><em>Hund's rule of maximum multiplicity</em></u>

Explanation:

<u><em>The aufbau principle:</em></u>

<em></em>

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

<u><em>The Pauli exclusion principle:</em></u>

<em></em>

Two electrons of the same atom cannot have their four equal quantum numbers.  Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

<u><em>Hund's rule of maximum multiplicity:</em></u>

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible.  Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy  the same orbital.

4 0
3 years ago
an object of mass 4 kg is placed on an inclined plane at an angle of 60. if the object slides down the plane with an acceleratio
balandron [24]

The forces on the  y axis are:

N-mgcos(60)=0   , wich becomes

N=mgcos(60)

Rember that the friction force is always contrary to the motion of an object and its formula is  f=μ * N

The forces in the x axis are:

-f + mgsin(60)= m * a

-μ*mgcos(60) + mgsin(60)=m*a  ,

μ = ( m*a - mgf=μ[sin(60) )/ ( mgcos(90) )


6 0
3 years ago
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