Ask question

Ask question

All categories

3 years ago

8

The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco

nds?

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer:

160 W

Explanation:

Power is the ratio of work to time:

(1600 J)/(10 s) = 160 J/s = 160 W

You might be interested in

Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please) :D

kap26 [50]

12.could be D
13.A
14.D
15.C
16.A

3 0

3 years ago

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

In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of

antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, $a=84\ au= 1.496\times 10^{11}\ m$

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e $T^2=\dfrac{4\pi^2}{GM}a^3$

where G is gravitational constant

M is mass of sun, $M=1.98\times 10^{30}\ Kg$

So, $T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}$

$T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3$

$T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}$

$T^2=59532469.8\times 10^{14}\ s$

$T=7715.7\times 10^7\ s$

since, $1\ sec=3.17\times 10^{-8}\ years$

So, orbital period is approximately 2446 years.

7 0

4 years ago

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings

Hatshy [7]

Answer:0.316 rad/s

Explanation:

Given

mass of Person $m=65 kg$

velocity of person $v=3.8 m/s$

diameter of turntable $d=6.5 m$

moment of Inertia of the table $I_0=1850 kg-m^2$

Moment of inertia of Person I

$I=mr^2=65\times (\frac{6.5}{2})^2$

$I=686.56 kg-m^2$

initial angular velocity $\omega _1=\frac{v}{r}$

$\omega _1=\frac{3.8}{3.25}=1.17 rad/s$

Conserving Angular momentum

$I\omega _1=(I+I_0)\omega _2$ , where $\omega _2=final\ angular\ velocity$

$686.56\times 1.17=(686.56+1850)\times \omega _2$

$\omega _2=\frac{686.56}{2536.56}\times 1.17$

$\omega _2=0.316 rad/s$

4 0

3 years ago

Acceleration can be found by computing the slope of a blank vs .time graph

scoundrel [369]

Velocity-time graph

5 0

3 years ago

Read 2 more answers