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zavuch27 [327]
3 years ago
11

you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

meter of the nozzle you hold is 5.45x10^-3 m. at what speed does the water exit the nozzle?
Physics
1 answer:
Ivanshal [37]3 years ago
7 0

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m

So its area must be

A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s

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A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 0
4 years ago
Please help, and show steps. Thank you very much!
Vikentia [17]
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
6 0
4 years ago
A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
Schach [20]

Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

<h3>a = 0.04 m s²</h3>
5 0
2 years ago
A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the
Anni [7]

Answer:

I=38.181\ A is the current through the body of the man.

E=34.5\ J energy dissipated.

Explanation:

Given:

  • time for which the current lasted, t=43\times 10^{-6}\ s
  • potential difference between the feet, V=21000\ V
  • resistance between the feet, R=550\ \Omega

<u>Now, from the Ohm's law we have:</u>

I=\frac{V}{R}

I=\frac{21000}{550}

I=38.181\ A is the current through the body of the man.

<u>Energy dissipated in the body:</u>

E=I^2.R.t

E=38.181^2\times 550\times 43\times 10^{-6}

E=34.5\ J

5 0
4 years ago
A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/
kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 16.0 kg is the mass of the canoe

u_1 = -12.5 m/s is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

v_1 is the final velocity of the canoe

m_2 = 14.0 kg is the mass of the raft

u_2 = +16.0 m/s is the initial velocity of the raft

v_2 = -14.4 m/s is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0
3 years ago
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