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natali 33 [55]
1 year ago
9

A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the

radius of Earth and is equal to 6.4×106m. What must be the radius of orbit for the same satellite in order for it to maintain a circular orbit and travel at a speed of 2v0 ?
Physics
1 answer:
NISA [10]1 year ago
8 0

Answer:

1.024 × 10⁸ m

Explanation:

The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.

So, ω =  v₀/8RE

For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have

2v₀ = Rω

substituting ω =  v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

2 = R/8RE

So, R = 2 × 8RE

R = 16RE

substituting RE = 6.4 × 10⁶ m

R = 16RE

= 16 × 6.4 × 10⁶ m

= 102.4 × 10⁶ m

= 1.024 × 10⁸ m

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a torch bulb is rated 2.5V and 750mA. Calculate its power,its resistance and the energy consumed if this bulb lighted for 4 hour
Hatshy [7]
Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
Substituting the values to the given, 
P = Power = ?
R = Resistance = ?
V = Voltage = 2.5 V
I = Current = 750 mA

R = V/I = 2.5/ (750 x 10^-3)
R = 3.33 ohms

Calculating the power, we have P = IV

P = (750 x 10^-3)(2.5) 
P = 1.875 W

The power consumption is the power consumed multiply by the number of hours. In here, we have;
1.875W x 4 hours = 7.5 watt-hours
3 0
1 year ago
A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward
Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

                            =\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)

                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

                                  ax=(change in velocity)/(Time)

                                  ax=\frac{-0.5-0}{3} =-\frac{1}{6} ms2

3 0
1 year ago
A microwave oven operates with sinusoidal microwaves at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the
Degger [83]

Answer:

F = 2 × 10⁻³ N

Explanation:

Given:

frequency, f = 2400 MHz

Height, h = 25cm = 0.25 m

Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²

Energy of the  microwave, E = 0.50 mJ = 0.5 x 10⁻³ J

Now, the time taken by the wave from top to the base, t = h/c

here, c is the speed of the light

thus,

t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s

The radiation pressure P_r = Intensity/c

now, the intensity is given as:

I = Power/ area

also,

Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W

thus,

I = 600000 W/ 0.09 m² = 6666666.6 W/m²

substituting the value in the formula for pressure due to radiation, we have

P_r = 6666666.6 W/m²/(3 x 10⁸)

also

pressure = Force/ area

thus,

Force/ area = 6666666.6 W/m²/(3 x 10⁸)

or

Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)

or

F = 2 × 10⁻³ N

6 0
1 year ago
Read 2 more answers
To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a
Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0
2 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
2 years ago
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