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3 years ago

10

Can someone help me with this. I'm not really sure if the right answer is c.

1 answer:

taurus [48]3 years ago

6 0

It might be c I'm not sure either. Hope I helped

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The center of mass of a 0.30-kg (non-uniform) meter stick is located at its 45-cm mark. What is the magnitude of the torque (in

lbvjy [14]

Answer:

The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

Explanation:

Given that,

Mass of the meter stick, m = 0.3 kg

Center of mass is located at its 45 cm mark.

We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

$\tau=r\times F\\\\\tau=(45-28)\times 10^{-2}\times 0.3\times 9.79\\\\\tau=0.499\ N-m$

or

$\tau=0.5\ N-m$

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

6 0

3 years ago

Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the

marin [14]

Answer:

Explanation:

Given

Position of squirrel is given by

$s(t)=t^3-12t^2+36t$

Velocity is given by

$v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}$

$v(t)=3t^2-12\times 2t+36$

$v(t)=3t^2-24t+36$

(b) acceleration is given by

$a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}$

$a(t)=6t-24$

(c)at $s(3)=3^3-12(3)^2+36(3)$

$s(3)=27\ m$

at $s(4)=4^3-12(4)^2+36(4)$

$s(4)=16\ m$

at $t=3\ s$ Position is $27\ m$ and at $t=4\ s$ position is $16\ m$

therefore squirrel is moving down

6 0

4 years ago

If the radius of a coin is 1 cm then calculate its area.

igor_vitrenko [27]

Answer:

3.14*1²

3.14 cm²

I hope this will help

5 0

3 years ago

Read 2 more answers

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago

What is the equation of the line that is perpendicular to y= -4x +5 and passes through the point (4,-3)?

Ostrovityanka [42]

Y=4x-19

Perpendicular because it has an opposite reciprocal slope

5 0

3 years ago