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Svetradugi [14.3K]
3 years ago
15

Janet jumps off a high diving platform with a horizontal velocity of 2.8 meters per second in lands in the_________.

Physics
1 answer:
Inessa05 [86]3 years ago
6 0

Answer:

Janet jumps off a high diving platform with a horizontal velocity of 2.89 m per s and lands in the water 1.5 s later. How high is the platform?

Platform is 11.025 meters high .

Explanation:

we have Vx = 2.89 m/s

time taken = 1.5 seconds

height of the platform = ?

so,

As Janet is jumping from a high diving platform from a certain unknown height their must be involvement of gravity in action.

we can use,

h = Vi*t+(1/2)*g*t^2

where ,

h = height

Vi = initial horizontal velocity that will be zero

t = time in seconds

g = gravity due to acceleration

now put the values

h = 0+(1/2)*(9.8)*(1.5)^2

h = 11.025-m

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A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?
Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force F between two charged objects:

\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}},

where

  • k is coulomb's constant. k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2} in vacuum.
  • q_1 and q_2 are the signed charge of the objects.
  • r is the distance between the two objects.

For the two electrons:

  • q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}.
  • r = 1\times 10^{-10}\;\text{m}.
  • \displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}.

The sign of F is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

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xxTIMURxx [149]
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On the Earth, insolation (We) = Psun/Ae

Therefore,
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6 0
3 years ago
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