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Jobisdone [24]
2 years ago
7

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m ​ 3, po

int, 7, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction downward. After 3\,\text{s}3s3, space, s, julia is falling downward with a velocity of 3.1\,\dfrac{\text{m}}{\text{s}}3.1 s m ​ 3, point, 1, space, start fraction, m, divided by, s, end fraction.
Physics
1 answer:
liberstina [14]2 years ago
6 0

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/s^2 . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s

So jumping velocity of Julia = 8 m/s

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One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the
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<span>Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20</span>
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3 years ago
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gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

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3 0
2 years ago
An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which
Irina-Kira [14]

Answer:

2\times 10^{-3}\ C

6000

1.2 J

3.33\times 10^{-6}\ F

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V = Voltage = 100 mV

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Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C

The charge flowing through the electrocytes in that amount of time is 2\times 10^{-3}\ C

The maximum potential is given by

V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000

The number of electrolytes is 6000

Energy is given by

E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J

The energy released when the electric eel delivers a shock is 1.2 J

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C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F

The equivalent capacitance of all the electrocyte cells in the electric eel is 3.33\times 10^{-6}\ F

8 0
2 years ago
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v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
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sergij07 [2.7K]

The two flaws in her experiment’s design are

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