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1 year ago

What is the difference between a star and a planet.

1 answer:

5 0

Answer: In the planet there is life and the star is a store of energy where the planet gets its light, life and energy.

Explanation:

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Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speake

Serga [27]

Answer:

Explanation:

Anyone could be leaning forward toward the speaker but be distracted and I believe if you're paying attention to the speaker, you would ask questions to make sure you're understanding what they are speaking

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2 years ago

Read 2 more answers

A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400

slava [35]

Answer:

28,699m

Explanation:

The force to make the box move should be μs.N=μs.m.g=m.|a|

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

$x = v \times t + \frac{1}{2} \times a \times {t}^{2}$

as the truck is stopping, the acceleration is negative. then,

$x = v \times t - \frac{1}{2} \times |a| \times {t}^{2}$

$|a| = v \div t \\ t = v \div |a|$

$x = v \times (v \div |a| ) - \frac{1}{2} \times |a| \times {(v \div |a|)}^{2}$

$x = v \times (v \div μs.g ) - \frac{1}{2} \times |a| \times {(v \div μs.g)}^{2}$

$x = \frac{{v}^{2}}{μs \times g} - \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} } = 28.699m$

28,699m

7 0

2 years ago

List three devices that use electric current from batteries and three that use regular house current.

Bond [772]

Charger
counsels
TV

Fan
light
car

4 0

3 years ago

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

Bezzdna [24]

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

a) Determine the angular accelerations of B1 and B2 when couple is applied

at point P ; Co = I* ∝B2'

150 = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 = 180 rad/sec

b) Determine the force exerted on B2 at P

T2 = mB1g + T1 -------- ( 1 )

where ; T1 = mB2g ( at point p )

= 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

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2 years ago

What do wind and moving water have in common? Select one: They both have electrical energy They both have elastic potential ener

ozzi

They both have kinetic energy because they are both moving and have force.
Hope this helped! :)

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2 years ago