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iris [78.8K]
3 years ago
11

Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.625×10−3 m3/s and

the diameter of the nozzle you hold is 5.19×10−3 m. At what speed v does the water exit the nozzle?
Physics
1 answer:
Tems11 [23]3 years ago
4 0

Answer:

<em>0.153 m/s</em>

<em></em>

Explanation:

The flowrate Q = 0.625 x 10-3 m^3-/s

The diameter of the nozzle d = 5.19 x 10^-3 m

the velocity V = ?

The cross-sectional area of the flow A = \pi d^{2}/4

==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2

From the continuity equation,

Q = AV

V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = <em>0.153 m/s</em>

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A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
3 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}kg/m × (11-x)m =  3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0
3 years ago
A toy gun uses a spring to project a 5.9-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
LekaFEV [45]

Answer:

Explanation:

Stored energy in spring = 1/2 k x² , k is spring constant , x is compression.

= 1/2 x 8 x (5.7 x 10⁻²)²

= 129.96 x 10⁻⁴ J

Energy lost due to friction = force x distance

= .035 x .17

= .00595 J

Energy used in providing kinetic energy to projectile.  

129.96 x 10⁻⁴  - .00595

.012996 - .00595

= .007046 J

So

1/2 m v² = .007046

v² = .007046  x 2 / .0059

= 2.3885

v = 1.545 m /s

8 0
3 years ago
Read 2 more answers
Two atoms each contain 6 protons, but one has 6 neutrons and the other has 8 neutrons. What are these atoms called? A. ions B. r
ki77a [65]

Answer:

These atomos are called isotopes.

Explanation:

Each chemical element is characterized by the number of protons in its nucleus, which is called the atomic number (Z).

The number of neutrons in the nucleus can vary. There are almost always as many or more neutrons than protons. The atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element. That is, isotopes are atoms whose atomic nuclei have the same number of protons but different numbers of neutrons.

So, <u><em>these atomos are called isotopes.</em></u>

3 0
3 years ago
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