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Luden [163]
3 years ago
11

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati

onal force on the platter is 5.0 N, what is the magnitude of the force parallel to the tray that tends to cause the platter to slide down the tray? (Disregard friction.)

Physics
1 answer:
kirill115 [55]3 years ago
5 0

Answer:1.039 N

Explanation:

Given

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta is parallel to the tray =1.039 N

5\cos \theta is perpendicular to the tray =4.890 N

You might be interested in
A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10
Rina8888 [55]

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

3 0
3 years ago
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its
Scilla [17]
<h2>Power of cheetah is 5576.85 W = 7.48 hp</h2>

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44  ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

7 0
3 years ago
A 30-kg child sits at the top of a 3-meter slide. After sliding down, the child is traveling 4 m/s. How much PE does he start wi
Jobisdone [24]
At the top:

         Potential Energy = (mass) x (gravity) x (height)

                                       = (30 kg) x (9.8 m/s²) x (3 meters)

                                       =      882 joules

At the bottom:

           Kinetic Energy  =  (1/2) x (mass) x (speed)²

                                       = (1/2) x (30 kg) x (3 m/s)²

                                       =        (15 kg)  x  (9 m²/s²)

                                       =              135 joules .

He had  882 joules of potential energy at the top,
but only  135 joules of kinetic energy at the bottom.

Friction stole  (882 - 135) = 747 joules of his energy while he slid down.
The seat of his jeans must be pretty warm.
6 0
3 years ago
In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of fric
lorasvet [3.4K]

Answer:

B. The maximum angle decreases

Explanation:

If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .

tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .

So the tan of angle of repose θ is proportional to coefficient of static friction.

If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie,  it at lower angle of inclination , the object will start slipping .

7 0
3 years ago
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Rudik [331]

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
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