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ryzh [129]
2 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]2 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]2 years ago
3 0
Frog hopper B❤️recent emojis
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Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the
Fofino [41]

Answer:

d = 2,042 10-3 m

Explanation:

The laser diffracts in the circular slit, so the process equation is

      d sin θ= m λ

The first diffraction minimum occurs for m = 1

We can use trigonometry in the mirror

        tan θ = Y / L

Where L is the distance from the Moon to Earth

Since the angle is extremely small

           tan θ = sin θ / cos θ

           Cos θ = 1

           tant θ = sin θ = y / L

We replace

           d y / L = λ

           d = λ L / y

Let's calculate

           d = 532 10⁻⁹ 3.84 10⁶/1 10³

           d = 2,042 10-3 m

5 0
2 years ago
Which situation is a result of human activities
Mila [183]
I would answer but you have no choices lol

7 0
3 years ago
A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi
Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0
3 years ago
The engineer of a train traveling at 30 m/sec sees a cow on the tracks. He applies the brakes and causes the train to accelerate
sineoko [7]

Answer:The train travels 105 meters after applying the brakes

Explanation:If he decelerates 1.5 every minute, then he went from 28,5 m/s, to 27.0 m/s, to 25.5 m/s, to 24.0 m/s, after 4 seconds. Add all this together and youll get 105 meters moved in 4 seconds after he hit the brakes, I dont have a notebook on me though sorry :/

8 0
2 years ago
A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter
expeople1 [14]

Answer: 56.87m/s^{2}

Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

F_{net}=F_{up}-W  (1)

Where:

F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

W=m.g=(3kg)(9.79m/s^{2}) (2)

W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

F_{net}=m.a  (6)

Finding the acceleration a:

a=\frac{F_{net}}{m}  (7)

a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

3 0
2 years ago
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