All categories

3 years ago

Please help me!!!!!!!!!

Physics

1 answer:

Nadusha1986 [10]3 years ago

4 0

Hi there! :)

$\large\boxed{17.32 m/s}$

Use the following equation in solving for kinetic energy:

KE = 1/2mv² where:

KE = kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

12,000 = 40v²

Divide both sides by 40:

12,000 / 40 = v²

300 = v²

Take the square root of both sides:

√300 = v

v ≈ 17.32 m/s

You might be interested in

Would You Rather...Carry a 50 lb box up 3 flights of stairs or a 25 lb box up 5 flights of stairs? Whichever choice you make, ju

Oksanka [162]

I would carry a 25 pound box up 5 flights, because it is lighter. You will wear out faster carrying a 50 pound box up 3 flights.

8 0

3 years ago

The distance versus this plot for a particular object shows a quadratic relationship. Which column of distance data is possible

SOVA2 [1]

Answer:

Explanation:

In Column A, the pattern is 1², 2², 3², etc. So x(t) = t², which is quadratic.

In Column B, the pattern is 2*1, 2*2, 2*3, etc. So x(t) = 2t, which is linear.

In Column C, the pattern is 9*1, 9*2, 9*3, etc. So x(t) = 9t, which is linear.

In Column D, the pattern is 1/1, 1/2, 1/3, etc. So x(t) = 1/t, which is reciprocal.

In Column E, the pattern is 1/1², 1/2², 1/3², etc. So x(t) = 1/t², which is reciprocal.

5 0

4 years ago

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

3 years ago

I need help plz and thank you this is due

xeze [42]

Answer:

Grow up man, this is completely based on your curriculum, we would need your book to answer, and this has to be done by you.

3 0

3 years ago

Read 2 more answers

GIVING BRAINLIEST TO THE CORRECT ANSWER!!

Basile [38]

The answer is A. locations by the ocean typically do not get as cold in the winter or as hot in the summer as locations that are located inland.

6 0

3 years ago

Read 2 more answers