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k0ka [10]
3 years ago
13

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

Physics
1 answer:
just olya [345]3 years ago
6 0
Force is defined as the rate of change of momentum.
The initial amount of momentum is mv because water stops when it hit the wall total change of momentum must be \Delta p=mv.
Now let's calculate the force.
F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v
We need to find \frac{dm}{dt}. This is the amount of water hiting the wall per second.
\frac{dm}{dt}=\rho Av
Our final formula would be:
F=\rho Avv=\rho Av^2
And now we can calculate the answer:
F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N


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The volume of a piece of metal of a mass 6 gram is 15cm3. What is the density of the metal piece?
torisob [31]

Answer:

0.4 g/cm^3

Explanation:

The density of an object can be found using the following formula.

d= m/v

where m is the mass and v is the volume.

The mass of the metal is 6 grams and the volume is 15 centimeters^3

m=6 g

v= 15 cm^3

Substitute these into the formula.

d= 6 g/ 15 cm^3

Divide 6 g by 15 cm^3 (6/15=0.4)

d= 0.4 g/ cm^3

The density of the metal is 0.4 grams per cubic centimeter.

4 0
3 years ago
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A sound can be ________ or _________.
marin [14]
A.
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3 0
3 years ago
A swimmer moves through the water at 2.5 meters per second and experiences a drag force of 240 N. How much power is she generati
mote1985 [20]
-- There's a force of 240N pushing her backwards.

-- She's maintaining a steady speed (of 2.5 m/s) .

-- In order to maintain a steady speed (no acceleration),
the forces on her must be balanced.  So she's maintaining
a steady force of 240N forward.

-- Every time she moves 1 m forward, she does work of
(force) x (distance) = 240 joules.

-- She moves 2.5 meters forward every second.
So she's doing (240 x 2.5) = 600 joules of work every second.

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6 0
3 years ago
A metal ball has a net charge of 4.5x10-7 C
netineya [11]

a) the number of protons is 2.81\cdot 10^{12} more than the electrons

b) 4.69\cdot 10^{-15} kg

Explanation:

The net electric charge on the ball is

Q=+4.5\cdot 10^{-7}C

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.

The charge of one proton is:

q_p =+e= +1.6\cdot 10^{-19}C

While the charge of one electron is

q_e = -e=-1.6\cdot 10^{-19}C

So the net charge on the metal ball will be given by

Q=N_p q_p + N_e q_e = (N_p -N_e)e

where

N_p is the number of protons

N_e is the number of electrons

So we find:

N_p-N_e=\frac{Q}{e}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}

This means that the number of protons is 2.81\cdot 10^{12} more than the electrons.

b)

In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

Q-Q'=0

This implies that the charge that we must remove is

Q'=Q=4.5\cdot 10^{-7}C

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.

The number of protons that must be removed is:

N_p = \frac{Q'}{q_p}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}

The mass of one proton is

m_p = 1.67\cdot 10^{-27}kg

Therefore, the total mass that must be removed from the ball is

M=m_p N_p = (1.67\cdot 10^{-27})(2.81\cdot 10^{12})=4.69\cdot 10^{-15} kg

6 0
3 years ago
A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.
kakasveta [241]
The first law of thermodynamics says that the variation of internal energy of a system is given by:
\Delta U = Q + W
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system: 
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
\Delta U = -1275 J+855 J=-420 J
6 0
3 years ago
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