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3 years ago

7

If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass

1 answer:

EleoNora [17]3 years ago

5 0

Simply, apply the formula $E = \frac{1}{2}mv^{2}$ and insert the values of m = mass, v = velocity and E = Energy.
The result will be $72J = \frac{1}{2}m(12m/s)^{2}$ , m = 1 kg

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List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.

frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

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2 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

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4 0

3 years ago

Can someone please help me

guajiro [1.7K]

ANSWER:
C. Small, minimize

Hope it helps u!

5 0

3 years ago

A projectile is shot on level ground with a

erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above

5 0

2 years ago

a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp

stiks02 [169]

Answer:

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This is the KE due to the pendulum falling from a 25 deg displacement

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v = 1.92 m/s this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

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1 year ago