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2 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Physics

1 answer:

Degger [83]2 years ago

7 0

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

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A 326 g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the

Basile [38]

Answer:

a) $v_{max}=5.98\frac{m}{s}$

b) $k=205.92\frac{N}{m}$

c) $A=0.24m$

Explanation:

a) In the equilibrium position of the system, that is when the spring is not elongated, the potential energy is zero. Therefore, the total energy of the system is the maximum kinetic energy:

$E=K_{max}\\E=\frac{mv_{max}^2}{2}\\v_{max}=\sqrt{\frac{2E}{m}}\\v_{max}=\sqrt{\frac{2(5.83J)}{0.326kg}}\\v_{max}=5.98\frac{m}{s}$

b) The force constant of the spring can be calculated from the natural frequency of the system:

$\omega^2=\frac{k}{m}\\k=m\omega^2$

Recall that $\omega=\frac{2\pi}{T}\\$ , that is the distance traveled in one revolution divided into the time of one revolution. Replacing and solving for k:

$k=\frac{m4\pi^2}{T^2}\\k=\frac{(0.326kg)4\pi^2}{(0.25s)^2}\\k=205.92\frac{N}{m}$

c) The maximum speed is directly proportional to the amplitude of the motion:

$v_{max}=A\omega\\A=\frac{v_{max}}{\omega}\\A=\frac{(v_{max})T}{2\pi}\\A=\frac{(5.98\frac{m}{s})0.25s}{2\pi}\\A=0.24m$

3 0

3 years ago

14. Saeed is pulling a 6 kg heavy rock with an upward force of 40 N but does not succeed to lift it up. What is the magnitude of

NemiM [27]

Answer:

58.8 N

Explanation:

The normal force is calculated as equal to the perpendicular component of the gravitational force.

Thus; N = mg

We are given m = 6 kg

Thus;

N = 6 × 9.8

N = 58.8 N

Thus, magnitude of normal force on the rock = 58.8 N

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2 years ago

What is Br charge after it gains an electron?

sergeinik [125]

Im not 100% sure but i think its bromine.

Hope this helps ^_^

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3 years ago

How much heat is absorbed by 60 g of copper when it is heated from 20°C to 80°C?

ira [324]

Answer:

1,836J

Explanation:

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3 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago