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lianna [129]
3 years ago
7

What are the horizontal and vertical components of a lizard’s displacement if it has climbed 7m directly up a tree?

Physics
1 answer:
madam [21]3 years ago
7 0

Answer:

The horizontal component is zero.

The vertical component is 7\sin\theta

Explanation:

Given that,

The lizard climb 7m directly up on a tree.

We know that,

The horizontal component is

x=\cos\theta

The vertical component is

y=\sin\theta

If the lizard climb 7m directly up on a tree then,

We need to find the components

Using given data

The horizontal component of lizard is

x=0

The vertical component is

y=7\sin\theta

Hence, The horizontal component is zero.

The vertical component is 7\sin\theta

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Answer:

0.07°C

Explanation:

<u>solution:</u>

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v(T)=1480+4(T-4°C)

<u>at 4°C the travel time is:</u>

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<u>5°C, the travel time is:</u>

t(5◦C) = ( 7600 × 103 m ) / (1484 m/s) = 5188.7 s

<u>one degree C corresponds to a ∆t of 14 s so temperature difference is:</u>

ΔT=1 s/14 s=0.07◦C

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3 years ago
Blood contains positive and negative ions and therefore is aconductor. A blood vessel, therefore, can be viewed as anelectrical
CaHeK987 [17]

Answer:

<h2>Magnetic field required for the given induced EMF is 1.41 T</h2>

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Q = \frac{\pi E d}{4B}

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Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

F=\frac{kq^{2}}{r^{2}}

Solving the equation for q, we have:

kq^{2}=Fr^{2}

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q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

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3 years ago
Explanation for question i &amp; ii. Thank you.
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wouldn't it be 1000 km/h considering the plane was travelling the same speed for over half the trip?

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