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ololo11 [35]
3 years ago
9

Who the queen of rap

Physics
2 answers:
Inessa [10]3 years ago
6 0

Answer:

Nicki or cardi b idek....

Flauer [41]3 years ago
3 0

Answer:

MEGAN THEEEE STALLIAN

Explanation:

LOL NEED POINTS

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_________ is an umbrella term that applies to location-aware technologies that allow devices to communicate securely with each o
ratelena [41]

Answer:Near-field communication

Explanation:

Near Field Communication is an umbrella term that applies to location-aware technologies that allow devices to communicate securely with each other over short distances. Near-field communication is a short-range wirelessconnectivity standard that uses magnetic field induction to enable communication between devices when they're touched together, or brought within a few centimeters to each other.It also specifies a way for the devices to establish a peer-to-peer network to exchange data. After the peer to peer network has been configured, another wireless communication technology, such as Bluetooth or Wi-Fi, can be used for longer-range communication or for transferring larger amounts of data.

3 0
3 years ago
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Where in a river would you find fish that are adapted to low light and low oxygen levels?
JulijaS [17]
No place in the river is characterized like this
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3 years ago
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Select the correct image.
vivado [14]

Answer:

1st one mountains

Explanation:

PLATO/EDMENTUM

6 0
2 years ago
What is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)<br> please someone help me
777dan777 [17]

Answer:

60mph=26.8224meters per second

Explanation:

4 0
3 years ago
An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char
ArbitrLikvidat [17]

Answer:

Part a)

W = 1.58 \times 10^{-20} J

Part b)

v = 1.86 \times 10^5 m/s

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = 4.00 \times 10^{-12} C/m^2

now the electric field is given as

E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}

E = 0.225 N/C

Now acceleration of an electron due to this electric field is given as

a = \frac{eE}{m}

a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}

a = 3.97 \times 10^{10}

Now work done on the electron due to this electric field

W = F.d

d = 0.470 - 0.03

d = 0.44 m

So work done is given as

W = (ma)(0.44)

W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)

W = 1.58 \times 10^{-20} J

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

W = \frac{1}{2}mv^2 - 0

1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2

v = 1.86 \times 10^5 m/s

7 0
3 years ago
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