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4 years ago

Please!! List three things that can affect solubility.

Physics

1 answer:

Dennis_Churaev [7]4 years ago

7 0

Nature of the solute and solvent, Temperature, and Pressure all affect solubility.

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What would the acceleration of a car be from a stoplight if it started at 0 m/s and reached a

trapecia [35]

Answer:

The acceleration of a car would be: $a=12.8$ m/s²

Explanation:

Given

Initial velocity = $v_1=0$ m/s

Final velocity = $v_f=100$ m/s

Time elapsed = $t = 7.8$ s

To determine

We need to determine the acceleration of a car.

We know that acceleration is basically the rate of change in velocity over time.

Thus,

We can determine the acceleration using the formula

$\:\:\:a=\:\frac{v_f-v_i}{t}$

where

$a$ is the acceleration

$v_1$ is the initial velocity

$v_f$ is the final velocity

$t$ is time elapsed

now substituting the values $v_1=0$ , $v_f=100$ , and $t = 7.8$ in the formula

$\:\:\:a=\:\frac{v_f-v_i}{t}$

$a=\:\frac{100-0}{7.8}$

$a=\frac{100}{7.8}$

$a=12.8$ m/s²

Therefore, the acceleration of a car would be: $a=12.8$ m/s²

3 0

3 years ago

you stretch a spring by a distance of 0.3 m. the spring has a spring constant of 440 n/m. when you release the spring, it snaps

Alina [70]

Answer:

19.8 J

Explanation:

According to the law of conservation of energy, the total mechanical energy of the spring (sum of kinetic energy and elastic potential energy) must be conserved:

$K_i + U_i = K_f + U_f$ (1)

where we have

$K_i$ is the initial kinetic energy of the spring, which is zero because the spring starts from rest (2)

$U_i$ is the elastic potential energy of the spring when it is fully stretched

$K_f$ is the kinetic energy of the spring when it reaches the natural length

$U_f$ is the elastic potential energy of the spring when it reaches its natural length, which is zero because the stretch in this case is zero (3)

$U_i = \frac{1}{2}k(\Delta x_i)^2$

where

k = 440 N/m is the spring constant

$\Delta x_i = 0.3 m$ is the initial stretching of the spring

Substituting,

$U_i = \frac{1}{2}(440)(0.3)^2=19.8 J$

And so using eq.(1) and keeping in mind (2) and (3) we find

$K_f= U_i = 19.8 J$

8 0

4 years ago

A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from

kogti [31]

Answer:

Explanation:

7 0

4 years ago

A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t

Vlada [557]

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

3 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

3 years ago