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stiks02 [169]
3 years ago
8

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'

s tongue accelerates at a remarkable 260 m/s2 for 20 ms, then travels at constant speed for another 30 ms. During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Physics
1 answer:
kramer3 years ago
4 0
Chameleon's tongue is more fast than thought. Its long sticky tongue moves at an amazing ballistic speed which lashes out unsuspecting insects and bugs. Now let us see how fast it is.

GIven:

acceleration of the chameleon's tongue- 260 m/s
2 for 20 ms
constant speed 30 ms
50 ms total time
1/20 of a second

solution:

<u>260</u> = <u>  n</u><u>  </u>
20       50
<u>20 n </u>= <u>13, 000</u>
20            20
n= 650 m/s
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3 years ago
Picture is down below Thank you!​
Murljashka [212]
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7 0
2 years ago
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What linear speed must an earth satellite have to be in a circular orbit at an altitude of 159 km?
IgorLugansk [536]
Gravitational acceleration, g = GM/r^2. Additionally, for a satellite in a circular orbit, g = v^2/r

Where, G = Gravitational constant, M = Mass of earth, r = distance from the center of the earth to the satellite, v = linear speed of the satellite.

Equating the two expressions;
v^2/r = GM/r^2
v = Sqrt (GM/r);
But GM = Constant = 398600.5 km^3/sec^2
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3 years ago
The rate an object is moving relative to a reference point is its
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speed.

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2 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
3 years ago
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