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Aliun [14]
3 years ago
14

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

ogs sprint away in different directions while Daniel stands still. From Daniel's point of view, Newton runs due North at 3.40 m/s, but from Pauli's point of view, Newton appears to be moving at 1.10 m/s due East. What must Pauli's velocity relative to Daniel be for this to be true? Express your answer in terms of the ???? ‑ and ???? ‑components if North is the +???? ‑direction and East is the +???? ‑direction.
Physics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

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Explanation:

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2 years ago
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Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25
dsp73
Here, 

height at failure, h1 = 525 m, 
upward acceleration, a = 2.25 m/s^2, 
velocity = v m/s, 
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2, 
<span>
SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence, 
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters 

b) 
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec 

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4 0
3 years ago
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A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
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Answer:

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Explanation:

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3 years ago
When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b
Nesterboy [21]

Answer:

Explanation:

Given

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Kinetic Energy is \frac{8}{9} of Total Energy

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P.E.=\frac{1}{4}kA^2

\frac{1}{2}kx^2=\frac{1}{4}kA^2

x=\frac{A}{\sqrt{2}}                  

7 0
3 years ago
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