Answer:
Hope this help you!!
Explanation:
Crust : The crust is the thinnest layer of the Earth. It has an average thickness of about 18 miles below land, and around 6 miles below the oceans. The crust is the layer that makes up the Earth's surface and it lies on top of a harder layer, called the mantle.
Mantle : The mantle is the mostly-solid bulk of Earth's interior. The mantle lies between Earth's dense, super-heated core and its thin outer layer, the crust. The mantle is about 1,802 miles thick, and makes up a whopping 84% of Earth's total volume
Outer Core : The outer core is the third layer of the Earth. It is the only liquid layer, and is mainly made up of the metals iron and nickel, as well as small amounts of other substances. The outer core is responsible for Earth's magnetic field. As Earth spins on its axis, the iron inside the liquid outer core moves around.
Inner Core : It's Almost The Size of the Moon. The Earth's inner core is surprisingly large, measuring 1,516 miles across. It's Mostly Made of Iron. It Spins Faster Than the Surface of the Earth. It Creates a Magnetic Field.
Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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Answer:
I'm just in jss2 but I read physics. this is what I think
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
