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2 years ago

What happens to a liquid beyond its certain temperature

1 answer:

7 0

Answer: what do you mean ''beyond certain temperature'', Do you mean ask in cooling the liquid down or heating the liquid up?

Explanation:

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In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second

lakkis [162]

If he keeps that pace he will be at the 34 yard line

7 0

2 years ago

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

4 0

3 years ago

Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons

yarga [219]

Answer:

$q\approx 6.6\cdot 10^{13}~electrons$

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

$\displaystyle F=k\frac{q^2}{d^2}$

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

$\displaystyle q=\sqrt{\frac{F}{k}}\cdot d$

Substituting values:

$\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1$

$q = 1.05\cdot 10^{-5}~c$

This charge corresponds to a number of electrons given by the elementary charge of the electron:

$q_e=1.6 \cdot 10^{-19}~c$

Thus, the charge of any of the spheres is:

$\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}$

$\mathbf{q\approx 6.6\cdot 10^{13}~electrons}$

5 0

2 years ago

A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when

vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

$v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}$

Then F = ma = 0.04 x 2500 = 100N

5 0

2 years ago

A surfer is moving at a constant velocity of 5.2 m/s north relative to a wave which is pushing him west at a constant velocity o

Vika [28.1K]

The speed and distances are directly proportional. Use ratios to solve for vertical y-distance. The ratio of x-distance west to y-distance north equals the x-velocity to y-velocity.

x/y = vx/vy
41/y = 8.6/5.2
41/y = 1.65
41/1.65 = y
24.8 m = y

5 0

2 years ago