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Ierofanga [76]
3 years ago
11

Answer this question please and you will can’t the brain list

Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

We know that velocity equals to difference in distance divided by difference in time.

1st period: velocity is increasing. Where v = (8-0)/(3-0) = 8/3 m/s.

2nd one: velocity decreases. Where

v=(5-8)/(4-3) = (-3) m/s.

3rd one: velocity is constant because no change in distance, so v=0.

4th one: velocity increases again. Where v=(10-5)/(8-6)= 5/2 m/s.

5th one: velocity is constant again, so v=0.

6th one: velocity decreases, where

v=(6-10)/(15-12)= (-4)/3 m/s.

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A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket
Nadusha1986 [10]

Answer:

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.10s later.

Explanation:

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket’s acceleration?

6 0
3 years ago
3.
weeeeeb [17]

Answer:

I.72m/s²

II.8m/s²

Explanation:

acceleration equal velocity² divided by length

8 0
3 years ago
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

4 0
3 years ago
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