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zavuch27 [327]
4 years ago
15

Which of the following situations represents a negative displacement? (Assume positive position is measured vertically upward al

ong a y-axis.) a. A cat stands on a tree limb. b. A cat jumps from the ground onto a tree limb. c. A cat jumps from a lower tree limb to a higher one. d. A cat jumps from a tree limb to the ground. Please select the best answer from the choices provided A B C D

Physics
2 answers:
GenaCL600 [577]4 years ago
8 0

Answer:

<u>Option-(D):</u>  "A cat jumps from a tree limb to the ground."

Explanation:

<u>Negative displacement(Assume positive position is measured vertically upward along a y-axis):</u>

The displacement can be visualized going from the initial point been supposed as the ground level. While, the limb is considered more elevated as compared to the ground level. So, going from the lower level to the higher position is considered as positive displacement while the movement from the higher position to the lower point is considered as negative displacement.

bekas [8.4K]4 years ago
6 0
A is wrong. There is no displacement. She's just sitting there.

C and B are both going up.

D is going down <<<==== answer.
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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t
kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel 40\times 2=80 miles

thus its position vector after two hours is

r_A=-80sin35\hat{i}+80cos35\hat{j}

similarly B travels with 20 mph and in 2 hours

=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}

Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

v_A=-40sin35\hat{i}+40cos35\hat{j}

Velocity of B

v_B=20sin80\hat{i}+20cos80\hat{j}

Velocity of A w.r.t B

v_{AB}=v_A-v_B

v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}

4 0
3 years ago
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3 years ago
It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su
Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

\epsilon = BvL sin \theta

where

B=5\cdot 10^{-5} T is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

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Substituting numbers into the formula, we find

\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV

8 0
3 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

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The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

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=>  v^2  =  \frac{GM }{r}

substituting this into the equation

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Now the gravitational force of the planet is mathematically represented as

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Where M is the mass of the planet and  m is the mass of the satellite

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     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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damaskus [11]
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Hope this helps!
8 0
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