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4 years ago

14

How is the answer C?

1 answer:

Lyrx [107]4 years ago

6 0

Explanation:

first the rocket is moving quicker and quicker so KE increases and momentum is the same as it would be then constant before the explosion

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Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in diffe

koban [17]

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

$v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}\\$

but the time just before the balls hit the ground is

$t=\frac{2v_{0}sin\alpha}{g}$

Hence we have

ball A

$v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball B

$v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball C

$v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

Hence, all three balls have the same speed just before hit the groug

vA=vB=vC

I hope this is useful for you

regards

5 0

3 years ago

The new memory-driven computer being developed by NASA and Hewlett Packard Enterprises is lightweight, hundreds of times faster

sladkih [1.3K]

It is true because the new memory driven developed by NASA

7 0

3 years ago

Read 2 more answers

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

KIM [24]

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>

6 0

3 years ago

006 (part 1 of 2) 10.0 points

Wittaler [7]

Snow balls can be fun. In the winter time, and in the Turkey season

8 0

3 years ago

An object is in simple harmonic motion. Find an equation for the motion given that the frequency is 3⁄π and at time t = 0, y = 0

Delicious77 [7]

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:

y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.

The data are: ν= 3π; y(t=0)=0 and y'(0)=6.

Firstly we know that 2πν=ω then ω=6*π^2

Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0

Besides y'(t)=6*π^2*A*cos (6*π^2*t)

y'(0)=6=6*π^2*A*cos (6*π^2*0)

6=6*π^2*A then A= 1/π^2

Finally the equation is:

y(t)= 1/π^2 sin(6*π^2*t)

3 0

3 years ago

Read 2 more answers