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Sindrei [870]
4 years ago
13

An object of mass 10.0 kg is released from the top of an inclined plane which makes an angle of inclination of 30.0o with the ho

rizontal. the object slides along the inclined plane. the questions refer to the instant when the object has traveled through a distance of 2.00 m measured along the slope. the coefficient of kinetic friction between the mass and the surface is 0.200. use g = 10.0 m/s2. how much work is done by gravity?

Physics
1 answer:
evablogger [386]4 years ago
6 0
We define work as the net force multiplied with the distance traveled. We already know distance to be 2 meters. Thus, all we need to know is the net force.

When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are three forces acting on the box: gravitational force pointing straight down, normal force perpendicular to the slope denoted as Fn and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (10)(10)(cos 30) = 86.6 N
∑Fx = μk*Fn - mgsin30° = Fnet
          (0.2)(86.6 N) - (10)(sin30°) = Fnet
Fnet = 12.32 Newtons

Thus,
W = Fnet*d
W = (12.32 N)(2 meters)\
W = 24.64 Joules

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Examine the scenario. Two neutral objects, a balloon and a sweater, are rubbed against each other. Which choice most accurately
Ne4ueva [31]

Electrons move from the sweater to the balloon. The sweater becomes positively charged, while the balloon becomes negatively charged.

Explanation:

  • Sweater is a conductive material, which means it readily gives away its electrons.
  • Consequently, when you rub a balloon on Sweater, this causes the electrons to move from the Sweater to the balloon's surface.
  • The rubbed part of the balloon now acquired  a negative charge. Objects made of rubber, such as the balloon, are basically electrical insulators, meaning that they resist electric charges flowing through them.
  • This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral after electrostatic process.
4 0
3 years ago
A manometer using oil (density 0.900 g/cm3) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases
Mila [183]

Answer:

Rise in level of fluid is 0.11 m

Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm

Solution:

As per the question:

Density of oil, \rho_{o} = 0.900\ g/cm^{3} = 900\ kg/m^{3}

Change in Pressure in the tank, \Delta P = 7.28\ mmHg

Density of the mercury, \rho_{m} = 13.6\ g/cm^{3} = 13600\ kg/m^{3}

Now,

To calculate the rise in the level of fluid inside the manometer:

We know that:

1 mmHg = 133.332 Pa

Thus

\Delta P = 7.28\ times 133.332 = 970.656\ Pa

Also,

\Delta P = \rho_{o} gh

where

g = acceleration due to gravity

h = height of the fluid level

970.656 = 900\times 9.8\times h

h = 0.11 m

Now, if mercury is used:

\Delta P = \rho_{m} gh

970.656 = 13600\times 9.8\times h

h = 0.00728 m = 7.28 mm

3 0
4 years ago
A dumped harmonic oscillator of a mass of 500 g has a period of 0.5 second. The amplitude of the oscillation is decreasing 2.0 %
inn [45]

Answer:

The answer is below

Explanation:

The amplitude decreases by 2%  during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:

y = abˣ; where x ad y are variables, a is the initial value and b is the factor.

Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:

a = 10 cm, b = 2% = 0.02.

Therefore:

y = 10(0.02)ˣ

The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:

y = 10(0.02)²⁵

y= 3.355 * 10⁻⁴² cm

The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm

7 0
3 years ago
An initially uncharged air-filled capacitor is connected to a 5.67-V charging source. As a result, 3.49 × 10-5 C of charge is tr
joja [24]

Answer:

V = 5.67 V

Q' =  2.09 x 10^-4 C

Explanation:

Potential difference, V = 5.67 V

Charge, Q = 3.49 x 10^-5 C

dielectric constant, K = 5.99

Capacitance of the capacitor

C = Q / V = 3.49 x 10^-5 / 5.67 = 6.15 x 10^-6 F

Now the dielectric is inserted, so,

C' = K x C = 5.99 x 6.15 x 10^-6 = 3.687 x 10^-5 F

After insertion of dielectric, as the battery is connected, so potential difference remains same.

Charge, Q' = C' x V = 3.687 x 10^-5 x 5.67 = 2.09 x 10^-4 C

8 0
3 years ago
100 POINTS. PLEASE PROVIDE EXPLANATION
maksim [4K]

Answer:

60 kg

80 kg

Explanation:

Work is equal to the change in energy.

W = ΔE = E − E₀

Let's start with block B.  The work done by the tension force is equal to the change in energy.  Initially, the block has potential energy.  Finally, the block has kinetic energy.

W = ΔE

FΔy = ½ mv² − mgh

T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)

T (-2.0 m) = m (-2 m²/s²)

T = m (1 m/s²)

Now let's look at block A.  The work done by tension and against friction is equal to the change in energy.  Initially, the block has no energy.  Finally, it has both kinetic and potential energy.

W = ΔE

Fd = ½ mv² + mgh − 0

(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)

(T − Nμ) (2.0 m) = 120 J

T − Nμ = 60 N

Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.

N = mg cos θ

N = (4.00 kg) (10 m/s²) (⅘)

N = 32 N

Substitute:

T − 32μ = 60 N

If μ = 0, then T = 60 N and m = 60 kg.

If μ = ⅝, then T = 80 N and m = 80 kg.

5 0
3 years ago
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