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Thepotemich [5.8K]

2 years ago

11

Eight books, each 4.6 cm thick and of mass 1.8 kg, lie on a flat table. How much work work is required to stack them on top of o

ne another?

1 answer:

pantera1 [17]2 years ago

3 0

Answer:

529.92Newton Meters

Explanation:

Work=force x displacement

8books in total

force is 1.8kg

dispplacement is 4.6cm

times both units by 8 and u get

force - 14,4

displacement - 36.8

now times boht together and u get 529.92nm

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When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t

Vaselesa [24]

Answer:

toward the normal

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

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4 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta

nikklg [1K]

When a river flows into an ocean, it slows down and deposits materials in its delta

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3 years ago

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

JulsSmile [24]

$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

7 0

3 years ago

Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force F, where r is the dista

Marrrta [24]

Answer:

W = - 5.01 10¹⁰ J

Explanation:

Work is defined by the expression

W = ∫ F.dr

Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product

W = ∫ F dr

W = G m₁m₂ ∫ 1 /r² dr

W = G m₁ m₂2(-1 / r)

We evaluate between the lower limits r = Re and upper r = ∞

W = G m₁m₂ (-1 / Re + 1 / ∞)

W = - G m₁ m₂ / Re

Let's calculate

W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶

W = - 5.01 10¹⁰ J

4 0

3 years ago