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zavuch27 [327]
2 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]2 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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An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
Who can help me for this question plz:(
Vanyuwa [196]
Force = mass x acceleration

15 = mass x 4

Mass = 15/4

Mass = 3.75 Kg
8 0
3 years ago
S Five particles with equal negative charges -q are placed symmetrically around a circle of radius R. Calculate the electric pot
Scrat [10]

The four distinct charges' combined potentials make up the potential in the square's center. The amount of the charge and the distance from the charge both affect the potential caused by a point charge.

Therefore, the center's total potential is V=4V1=ks4 q.

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

Due to the physical characteristic of electric charge, charged material experiences a force when it is exposed to an electromagnetic field. An object that has no net charge is said to be neutral. Classical electrodynamics is the name given to an earlier theory of the interactions of charged particles.

You can have positive or negative electric charges (commonly carried by protons and electrons respectively). opposing charges attract one another whereas similar charges repel one another.

To learn more about charge from the given link:

brainly.com/question/9194793

#SPJ4

6 0
1 year ago
You are not harmed by contact with a charged metal ball, even though its voltage may be very high. Is the reason similar to why
Jobisdone [24]

Answer:

.

Explanation:

.

6 0
3 years ago
Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp
shutvik [7]

Answer: 500 Watts

Explanation:

Power P is the speed with which work W is done. Its unit is Watts (W), being 1 W=\frac{1 Joule}{1 s}.

Power is mathematically expressed as:

P=\frac{W}{t} (1)

Where t is the time during which work W  is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm  ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (2)

In this case, we have the following data:

F=500 N

d=10 m

t=10 s

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

W=(500 N)(10 m) (3)

W=5000 J (4)

Substituting (4) in (1):

P=\frac{5000 J}{10 s} (5)

Finally:

P=500 W

3 0
2 years ago
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