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3 years ago

An object has a. Kinetic energy of 17j and a mass of 27 kg, how fast is the object moving?

Physics

1 answer:

kolezko [41]3 years ago

4 0

Answer:

1.1 m/s

Explanation:

W = (mv^2)/2

W = 17J

m= 27kg

v = $\sqrt{\frac{2W}{m} }$ = $\sqrt{\frac{2*17}{27} }$ ≈ 1.1 m/s

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A wooden cube with the mass of 1kg is placed on a frictionless plane that makes an angle of 30° with the floor.

Anni [7]

Answer:

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Explanation:

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2 years ago

The inner planets are DIFFERENT from the outer planets mainly because they are A) colder. B) larger. C) comprised of gas. D) com

Vaselesa [24]

The inner planets are not colder or larger than the outer ones,
and they're not comprised of gas.

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3 years ago

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The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

2 years ago

While goofing off at the ice skating rink, a student takes off her shoes and places each of them on the ice. Her friend, a hocke

3241004551 [841]

Answer:

The right shoe

Explanation:

Both shoes have the same speed.

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3 years ago

Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .

jek_recluse [69]

Explanation:

Given that,

Force, $F=((-8i)+6j)\ N$

Position of the particle, $r=(3i+4j)\ m$

(a) The toque on a particle about the origin is given by :

$\tau=F\times r$

$\tau=((-8i)+6j) \times (3i+4j)$

Taking the cross product of above two vectors, we get the value of torque as :

$\tau=(0+0-50k)\ N-m$

(b) Let $\theta$ is the angle between r and F. The angle between two vectors is given by :

$cos\theta=\dfrac{r.F}{|r|.|F|}$

$cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }$

$cos\theta=\dfrac{0}{50}$

$\theta=90^{\circ}$

6 0

3 years ago