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aalyn [17]
3 years ago
11

if person walks 10 muted used in 10 seconds 10 metre due east in next 10 seconds and 10 dual not in next 10 seconds then he has

dash velocity​
Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

answer is 10

Explanation:

10

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Dan is 50 years old, is beginning to look back at the successes and failures in his life, and hopes to use this reflection to de
Mice21 [21]

Answer:

C. Midlife Crisis.

Explanation:

3 0
3 years ago
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PEg=mgh For a mass of 60kg and 10m high
Anuta_ua [19.1K]
Assuming that the gravitational field strength is 10 N/kg, the sum would look like this:
PEg=mgh
=60×10×10
=6000J
However, you will need to check that you are not meant to be using 9.8 N/kg as the gravitational field strength.
3 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!
Nesterboy [21]

The correct answer is - It is the part of the ocean where new crusts are formed.

The place marked with A on the map is the place in the ocean where the new crust is formed. That is a place where there's a divergent plate boundary, or rather a place where the tectonic plates are moving away from one another. The gap and cracks left between them are easy target for the magma from the mantle to penetrate towards the surface. As the magma reaches the ocean floor it starts to cool off very quickly, creating new crust, and slowly making a very large underwater mountain range known as mid-ocean ridge.

7 0
4 years ago
Read 2 more answers
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
An open bed truck is moving at 20 m/s and comes to a stop. A 100 kg crate sitting in the back of the truck remains at rest in th
zzz [600]

The work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

<h3>Work done by friction to prevent the crate from moving</h3>

The work done is determined by applying the principle of conservation of energy.

W = K.E

W = ¹/₂mv²

where;

  • m is mass of crate
  • v is speed of the crate with respect to the truck = 20 m/s

W = ¹/₂(100)(20²)

W = 20,000 J

Thus, the work done was done by friction to keep the crate from moving as the truck came to a stop is 20,000 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

3 0
2 years ago
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