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WITCHER [35]
3 years ago
14

Is mixture an example of water

Physics
2 answers:
Evgesh-ka [11]3 years ago
7 0
No I don't believe so
BabaBlast [244]3 years ago
6 0
No it is not a mixture of water
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SOMEONE HELP ME PLEASE
mihalych1998 [28]

Explanation:

By the second law of Newton we get the relation

F = ma

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In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v
maw [93]

Answer:

True or False

Explanation:

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easy 50% chance you are right

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3 years ago
You are driving home from school steadily at for 180 km. It then begins to rain and you slow to You arrive home after driving 4.
nata0808 [166]

Answer:

Explanation:

Question is incomplete

Assuming the question you have asked is

You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.

given,

speed of 95 km/h for 180 km

due to rain

speed is reduced to 65 km/h

distance traveled in 4.5 hour

time taken to travel 180 km

d = s x t

t = \dfrac{180}{95}

     t = 1.9 hr

distance traveled in time, t' = 4.5-1.9 = 2.6 hr

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d' = s x t'

d' = 65 x 2.6

d'= 169 Km

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3 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

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2 years ago
What are each layer on an atom​
Zarrin [17]
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