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alexandr1967 [171]
3 years ago
6

Cell phones do not work everywhere. There are places where your cell phone cannot get a signal

Physics
2 answers:
Ilya [14]3 years ago
7 0

Answer:

D

Explanation:

cell phones must be close enough to a tower to send a receive electromagnetic waves

shusha [124]3 years ago
5 0
There are probably no strong signals
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Please help me with the question below the best answer with an explanation will get brainliest
lesya [120]

Answer:

B

Explanation:

i hope this is correct.

6 0
2 years ago
The game was played by swinging the big mallet down hard enough to cause the bell to ring. If it took a 44 newton force to ring
nata0808 [166]
I think C. 33 newtons
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3 years ago
At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops
jasenka [17]

Answer:

force = 11.33 kg-m/s^{2}

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_{1}=mU

P_{1}= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=\frac{\Delta P}{\Delta t}

F = \frac{P_{1}-P_{2}}{T}

Kgm/s^2

F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}[/tex]

8 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1
lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0
3 years ago
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