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3 years ago

If a box is not accelerating on a slope of 20 degrees, what is the net Force acting on the box?

Physics

1 answer:

never [62]3 years ago

3 0

Answer:

The net force is zero, if there is not external forces. The net force if the body is accelerating will be the product of the mass by the acceleration.

Explanation:

If the box does not accelerate, there are two possible options, either the box moves at constant velocity or does not receive any external force to alter its balance.

Newton's Second Law, also known as the Fundamental Law of dynamics, determines a proportional relationship between Force and variation of the amount of motion or linear moment of a body. In other words, The Force is directly proportional to the mass and acceleration of a body.

$F=m*a\\where:\\m= mass [kg]\\a=acceleration[m/s^2]$

$0=m*a\\a=0[m/s^2]$

If there are forces acting on the body and these forces are not in balance the body will move with a resulting force equal to the product of mass by acceleration.

$F_{net} =m*a$

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What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?

Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

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5 0

1 year ago

Party hearing. As the number of people at a party increases, you must raise your voice for a listener to hear you against the ba

dusya [7]

Answer:

$\frac{r_i}{1.77} m$

Explanation:

Given that

At starting separated = 1.20m

And the increase in background noise by Δβ = 5 dB, due to which the level of sound also rises

Based on the above information, the separation rf that is needed is shown below:

As we know that

$I_f = I_o \times 10^{\frac{\beta}{10} }\\\\I_f = I_i \times 10^{0.5}\\\\I_f = 3.16 \times I_i\\\\I \alpha \frac{1}{r^2} \\\\\frac{r_i^2}{r_f^2} = 3.16\\\\r_f = \frac{\sqrt{r_i^2}}{{\sqrt3.16}} \\\\= \frac{r_i}{1.77} m$

Hence, the separation r_f i.e. required is $\frac{r_i}{1.77} m$

We simply applied the above equation so that the correct separation could come

6 0

3 years ago

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared (units are in meters). If

ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$ ........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0

3 years ago

Two resistors are connected in parallel to a 12 V battery. The potential difference across one of the resistors is 12 v . Calcul

JulsSmile [24]

Answer:

See the explanation below.

Explanation:

We have to take into account that the potential difference is equal to the voltage, and this is measured between two points as the resistors are connected in parallel to the voltage source, the resistors will have the same voltage.

For ease, we will take the attached image of resistors connected in parallel.

As both resistors at their ends share the A & B connection points, these are at a voltage of 12V

5 0

3 years ago

A 0.89 kg ball is moving horizontally with a speed of 3.8 m/s when it strikes a vertical wall. The ball rebounds with a speed of

Julli [10]

Answer:

The change in momentum, p = -4.539 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 0.89 kg

Initial speed of the ball, u = 3.8 m/s

Final speed of the ball, v = -1.3 m/s (as it rebounds)

Let p is the change in linear momentum of the ball. We know that the linear momentum is equal to the product of mass and velocity. It is given by :

$p=m(v-u)$

$p=0.89\ kg\times ((-1.3)-3.8)\ m/s$

p = -4.539 kg-m/s

So, the change in linear momentum of the ball is 4.539 kg-m/s. Hence, this is the required solution.

4 0

3 years ago