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ryzh [129]
3 years ago
6

Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while

froghopper B jumps at a takeoff angle of 58o above the horizontal. Which froghopper experiences the greatest change in kinetic energy from the start of the jump to when it reaches the leaf?
Physics
2 answers:
IRINA_888 [86]3 years ago
6 0

Answer:

change in kinetic energy of both froghopper will be same

Explanation:

By work energy theorem we know that

Work done by all forces = change in kinetic energy

Here we know that two froghopper are sitting at same height and both jumps with same kinetic energy at different angles

1) First jump vertically upwards

2) Second jumps at an angle of 58 degree

so here while they are in air only gravitational force will work on it

so work done by gravity = change in kinetic energy

since both are initially at same height so work done by gravity on both of them will be same

So change in kinetic energy of both froghopper will be same

evablogger [386]3 years ago
3 0
Frog hopper B❤️recent emojis
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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:

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if you drop a softball from just above your knee, the kinetic energy of the ball just before it hits the ground is about 1 jule,
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The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is
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Answer and Explanation:

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