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cestrela7 [59]
3 years ago
9

PLEASE HELP!!!!

Physics
1 answer:
Alex_Xolod [135]3 years ago
5 0
I would say A Because it weighs more than the water
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A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f
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Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

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4 years ago
Which part of the electromagnetic specturm has the lowest radiant energy?
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The distance between two objects is increased by three times the oringinal distance. How will this change the force of attractio
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The prismatic bar has a cross-sectional area
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8 0
3 years ago
A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
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