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2 years ago

5

did anyone do the, kinetic energy lab report in physics?? plzzz help a girl outttttttt i need answerss????ill give you as much b

rainlist or points

1 answer:

Tema [17]2 years ago

8 0

Answer:

yes

Explanation:

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A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p

Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

$v_x = v cos45$

in y direction

$v_y = vsin45$

now displacement in x direction

$x = (vcos45)t + 0$

displacement in y direction

$y = (vsin45)t - \frac{1}{2}gt^2$

now from above two equations we have

$y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2$

$y = xtan45 - \frac{1}{2v^2cos^245}gx^2$

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0

3 years ago

How long does it take a 750-w coffeepot to bring to a boil 0.75 l of water initially at 11°c? assume that the part of the pot wh

Evgen [1.6K]

Just add all of them up and there is your answer I just added it but I want u to work it out to..

4 0

3 years ago

A 76 N crate is hung from a spring

lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0

2 years ago

If two equal charges are separated by a certain distance, the force of repulsion is F. Given F1, if each charge in F1 is doubled

Alisiya [41]

Answer: $F_2=16\times F_1$

Explanation:

Given

Force of repulsion between two charge particle is given by force F

Electrostatic force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ is the charges of particle

r=distance between charge particle

$=\frac{kq^2}{r^2}$ when charges are doubled and distance is reduced to half

i.e. q become 2 q and r becomes 0.5 r

$F_2=\frac{k(2q)^2}{(0.5r)^2}$

$F_2=\frac{kq^2}{r^2}\times 4\times 4$

$F_2=16\times F_1$

6 0

3 years ago

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum

wlad13 [49]

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

5 0

2 years ago