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2 years ago

Which of the following choices correctly pairs the organism with its phylum?

Physics

1 answer:

castortr0y [4]2 years ago

8 0

My lg is driplikemessiah and I don’t know which one it is but I say go with C

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Energy is the capacity to do what ? motion or work

Ber [7]

Answer:

energy is the capacity to do work

8 0

3 years ago

Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p

jekas [21]

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

7 0

3 years ago

How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2

olga_2 [115]

Golf no doing what I iuroeurir to do my homework and my homework

7 0

3 years ago

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A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +16.7 m/s in a time of 20.7 s.

frutty [35]

Answer:

F = 479.21 N

Explanation:

given,

initial velocity = 0 m/s

final velocity = 16.7 m/s

time taken = 20.7 s

combined mass of the boat and trailer = 594 kg

tension in the hitch = ?

using equation of motion

v = u + a t

16.7 = 0 + a × 20.7

a = 0.807 m/s²

Force = mass × acceleration

F = 594 × 0.807

F = 479.21 N

Hence, the tension in the hitch that connects the trailer to the car is F = 479.21 N

7 0

3 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers