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solniwko [45]
3 years ago
9

two womans are of the same weight. one wears sandals with pointed heels while the other wears sandals with flat sole s. which on

would feel more comfortable while walking on a sandy beach? give reasons to support your answer​
Physics
1 answer:
BlackZzzverrR [31]3 years ago
7 0

Answer:

flat sole

Explanation:

the pointed heels will sink in the sand whereas the flat sole will not.

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A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
hram777 [196]

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

4 0
2 years ago
If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth
fgiga [73]

<u>Answer:</u> The elevation in boiling point is 1.024°C.

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=ik_b\times m

where,

i = Van't Hoff factor = 2 (for NaCl)

\Delta T_b = change in boiling point  = ?

k_b = boiling point constant = 0.512^oC/m

m = molality = 1.0 m

Putting values in above equation, we get:

\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC

Hence, the elevation in boiling point is 1.024°C.

5 0
3 years ago
What is the formula you use to determine the gravitational potential energy of a object
kozerog [31]

Gravitational potential energy, relative to some level =

       (mass of the object)
times
       (height above the reference level)
times
       (acceleration due to gravity) .
4 0
3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
2 years ago
Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a
nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

    p₀o = m₁ v₁ + m₂ v₂

    pf = 0

    m₁ v₁ + m₂ v₂ = 0

    m₁ / m₂ = -v₂ / v₁

    m₁ / m₂=  - (-6.2) / 4.7

     m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

    Xcm = 1/M    (m₁ x₁ + m₂ x₂)

We derive from time

   Vcm = 1/M   (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

    0 = 1/M   (m₁ v₁ + m₂v₂)

   m₁ v₁ + m₂v₂ = 0

    m₁ / m₂ = -v₂ / v₁

   m₁ / m₂ = 1.3

4 0
3 years ago
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