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alexandr1967 [171]
3 years ago
12

420 hg = _____ cg help please

Physics
2 answers:
kondaur [170]3 years ago
4 0
4200000 is your answer hope this helps
Ivanshal [37]3 years ago
3 0
I got 4200000 for this one.
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I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i
Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

  = ½ (10 m/s2) (4 s)2

  = (5 m/s2) (16 s2)

  = 80 ms

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2 years ago
A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th
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Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

   \omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}

   \omega = 1.864\times 10^{-7}\ rad/s

b) tangential speed

   v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}

       v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

      a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}

          a = 3.62 x 10⁻³ m/s²

8 0
3 years ago
Two forces, F1 and F2, are applied to a block on a frictionless, horizontal surface
Komok [63]

Answer:

F, = 12N. F, = 2 N. Block. 4) a 20.0-kg mass moving at 1.00 m/s.

Explanation:

5 0
2 years ago
Read 2 more answers
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
DerKrebs [107]

Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

A)

  • Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.
  • So, we can apply the definition of average velocity to find this speed as follows:

       v_{x} = \frac{\Delta x}{\Delta t}  (1)

  • We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.
  • This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.
  • Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).
  • Since the acceleration is constant, we can use the following kinematic equation, as follows:

       \Delta y  = y_{f} - y_{o} = v_{o} * t + \frac{1}{2}  * g *t^{2}  (2)

  • if we take the river level as our x-axis, this means that yf = 1.3 m and

       y₀ = 20.8 m.

  • At the same time, due to in the vertical direction the car has no initial velocity, this means that  v₀ = 0.
  • Replacing by the values in (2) , and solving for t:

       t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} }  = 2 s  (3)

  • If we choose t₀ =0 ⇒ Δt = t = 2 s
  • Replacing Δx and Δt in (1):

       v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s  (4)

B)

  • When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.
  • Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

      v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s  (5)

  • Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

       v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}}  = 33.0 m/s  (6)

5 0
3 years ago
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