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Gennadij [26K]
3 years ago
7

The Thermosphere contains a layer of charged particles called the ________________ which makes commucation by _________________

possible and is home to the __________________ (Nothern Lights). This is the layer in which ____________________ fly.
Physics
1 answer:
fredd [130]3 years ago
7 0
Ionosphere
Radio wave transmission
Aurora Borealis
Airplanes
You might be interested in
A baseball pitcher throws a 0.14 kg ball toward a batter who is 18 m away.
Olin [163]

Answer:

A. 112 J

Explanation:

KE = ½mv² = ½(0.14)40² = 112 J

5 0
3 years ago
The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
2 years ago
1. A racing car with the driver weighs 1825 lb. Find the kinetic energy in ft*lb when traveling with a speed of 100 mi/hr.
Svetlanka [38]

Answer:

1. 610,000 lb ft

2. 490 J

Explanation:

1. First, convert mi/hr to ft/s:

100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s

Now find the kinetic energy:

KE = ½ mv²

KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²

KE = 610,000 lb ft

2. KE = ½ mv²

KE = ½ (5 kg) (14 m/s)²

KE = 490 J

6 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Why are moving iron instrument are of repulsion type
never [62]

Answer:

he spring provides the controlling torque. The air friction induces the damping torque, which opposes the movement of the coil. The repulsion type instrument is a non-polarized instrument, i.e., free from the direction of current passes through it. Thus, it is used for both AC and DC

5 0
3 years ago
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