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Diano4ka-milaya [45]
3 years ago
9

What is newton's first low of motion?​

Physics
2 answers:
sergey [27]3 years ago
8 0
Newton's first law of motion is that an object in motion will tend to stay in motion unless an external force acts upon it.
densk [106]3 years ago
7 0

Answer:

gravity is the newton first low of motion

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The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If t
My name is Ann [436]

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the  speed in larger pipe

\phi =\pi r^2 v

v = \frac{\phi}{\pi r^2}

  = \frac{2900 cm^3/s}{3.14(1.25cm)^2}

= 591.10 cm/s

 = 5.91 m/s

by Bernoulli's EQUATION

p1 +\frac{1}{2} \rho v1^2 = p2 +\frac{1}{2} \rho v2^2

139000+ \frac{1}{2}*1000*5.91^2 = 101000 +\frac{1}{2}*1000* v2^2

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

q = v \pi r^2

r^2 = \frac{q}{\pi v}

     = \frac{2900}{3.14*1053}

r = 0.87 cm

d = 2*0.87 = 1.75 cm

5 0
3 years ago
Physics help!! 10pt
Butoxors [25]
<span>A. accelerate toward the more massive object is your answer. In fact, even the acceleration with accelerate, also known as the jer.k. The more massive object will pull on the smaller object and the smaller object will increase its speed towards the more massive object.


idk why jer.k is a swear</span>
7 0
3 years ago
Read 2 more answers
a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?
Sedaia [141]

Explanation:

power =  \frac{energy \: expended}{time}

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

3 0
4 years ago
A student produces a wave in a long spring by
Sliva [168]

Answer: 2

Explanation:

4 0
3 years ago
Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0
3 years ago
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