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stiks02 [169]
3 years ago
6

Which is an example of a solution?

Physics
1 answer:
vladimir1956 [14]3 years ago
6 0

Answer:

C

Explanation:

plato

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Please help asap its for anatomy
Brilliant_brown [7]

The smooth muscle in the wall of the bladder when stretched triggers the micturition reflex (urination).

<h3>What is a Bladder?</h3>

This is defined as a lined layers of muscle tissue that stretch to hold urine in organisms.

In older people the elasticity of the bladder is reduced which is why it makes it harder for them to hold urine for a long time.

Read more about Micturition here brainly.com/question/26493943

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2 years ago
Determine the speed
Paraphin [41]
Calculating the average speed is simple using the formula <span>speed = distance/time</span>
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4 years ago
What force is required to give an object with mass 2000 kg an acceleration of 3.5 m/s2
ch4aika [34]
According to Newton's Second Law:
F = m*a = 2000 kg*3.5 m/s^2 = 7000 N
8 0
4 years ago
Read 2 more answers
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
3 years ago
Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th
kari74 [83]

When they meet the 40-kg boy will have moved a distance of 6 m.

Displacement of the 40 kg boy

The displacement of the 40 kg boy is calculated from the principle of center mass.

X(40 kg) = (60 x 10 m    +  40 x 0)/(40 kg + 60 kg)

X(40 kg) = (600)/(100) = 6 m

X(60 kg) = (60 x 0    +  40 x 10 m)/(40 kg + 60 kg)

X(60 kg) = (400)/(100) = 4 m

Thus, when they meet the 40-kg boy will have moved a distance of 6 m.

Learn more about center mass here: brainly.com/question/13499822

#SPJ1

7 0
2 years ago
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