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deff fn [24]
3 years ago
6

Hàng ngày ta thấy Mặt trời , Mặt trăng quay quanh Trái Đất, khi đó ta đã lấy vật mốc là

Physics
1 answer:
marshall27 [118]3 years ago
5 0
I’m sorry I don’t understand this language
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What are the units of energy?
MaRussiya [10]
The unit is Joule (J)
3 0
3 years ago
Read 2 more answers
a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

  Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock.  A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

    His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock.  The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so 

                     (300 x speed) = 36 kg-m/s .

Divide each side by 300:  speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed 
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
                                                         (300 x speed) = 186

Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

  yay !

By the way ... thanks for the 6 points.  The warm cloudy water
and crusty green bread are delicious.


4 0
3 years ago
If we cannot see the whole society what can we see
creativ13 [48]
I believe it's People interacting. I hope i've helped you! (:

8 0
3 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would
Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

W = mg

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

W_{x} = mg_{x}

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

34.65 N = (1.5 kg)g_{x}

g_{x} = \frac{34.65 N}{1.5 kg}

<u>gₓ = 23.1 m/s²</u>

7 0
3 years ago
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