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3 years ago

Which two concepts are Newton´s laws based on?

Physics

2 answers:

natita [175]3 years ago

7 0

Answer:

Inertia and Force.

sukhopar [10]3 years ago

6 0

Instantaneous impulse and continuous force: these are the foundations of Newton laws. In short explanation; concept of force and mass.

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A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250

Anettt [7]

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

= 6 hours

Total distance traveled = 350 km + 250 km

= 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

8 0

3 years ago

Water flows through a pipe of

snow_lady [41]

Answer:

9.38 m/s

Explanation:

Mass is conserved.

m₁ = m₂

ρ₁ Q₁ = ρ₂ Q₂

Assuming no change in density:

Q₁ = Q₂

v₁ A₁ = v₂ A₂

v₁ π r₁² = v₂ π r₂²

v₁ r₁² = v₂ r₂²

Plugging in values:

(1.50 m/s) (0.0250 m)² = v (0.0100 m)²

v = 9.38 m/s

3 0

3 years ago

You want to design an oval racetrack such that 3200 lb racecars can round the turns of radius 1000 ft at 1.00 _ 102 mi/h without

Svetradugi [14.3K]

<span>v = (22/15)*98 = 143.7 ft/sec
For no friction, Θ = arctan[v²/(R*g)] = 32.68°

Anet = √[(v²/R + g²] = 33.2 ft/sec²
Af = Anet*sinΘ = 17.93 ft/sec²
Ff = m*Af = (3200/32.2)*Af = 1781.9 lb along the slope.
Fr = Ff*cosΘ = 1500 lb</span>Source(s):<span>Free body diagram
My 'cheat sheet' of physics formulae</span>

3 0

3 years ago

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 12.0 m/s. how long does h

Lunna [17]

The shot has a constant downward acceleration of $a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}$ . Since acceleration is constant, the velocity of the ball is given by

$a_y=\dfrac{v_y-v_{0y}}t$

where $v_y$ is the shot's vertical velocity at time $t$ and $v_{0y}$ is its initial vertical velocity. We're given that $v_{0y}=12.0\,\dfrac{\mathrm m}{\mathrm s}$ . So

$-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-12.0\,\frac{\mathrm m}{\mathrm s}}t\implies v_y=12.0\,\dfrac{\mathrm m}{\mathrm s}+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t$

gives the shot's velocity at time $t$ .

Because acceleration is constant, we also know that

$\bar v_y=\dfrac{d_y-d_{0y}}t=\dfrac{v_y+v_{0y}}2$

where $\bar v_y$ is the average vertical velocity, $d_y$ is the vertical displacement at time $t$ , and $d_{0y}$ is the initial displacement at $t=0$ . We're given an initial displacement of $d_{0y}=2.10\,\mathrm m$ , so the displacement of the shot is

$d_y=d_{0y}+\dfrac12(v_y+v_{0y})t=d_{0y}+v_{0y}t+\dfrac12a_yt^2$

$\implies d_y=2.10\,\mathrm m+\left(12.0\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

The shotputter is 1.90 m tall, a difference of 0.20 m from the initial height of the throw. So we want to find the time for the ball to reach a displacment of -0.20 m:

$-0.20\,\mathrm m=2.10\,\mathrm m+\left(12.0\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$