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choli [55]
3 years ago
13

A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much wo

rk does the cannon do on the cannonball? (b) The cannon is aimed at a 25.0° angle above the horizontal. Assume gravity is the only other force acting on the cannonball as it moves through the cannon barrel. (That is, ignore all frictional forces.) What is the net work done by these 2 forces on the cannonball while it is in the cannon barrel?
Physics
1 answer:
Zarrin [17]3 years ago
3 0

To develop this problem it is necessary to apply the concepts related to Work and energy conservation.

By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,

W = F*d

Where,

F= Force

d = Distance

On the other hand we know that the potential energy of a body is given based on height and weight, that is

PE = -mgh

The total work done would be given by the conservation and sum of these energies, that is to say

W_{net} = W+PE

PART A) Applying the work formula,

W = F*d\\W = 2750*1.25\\W = 3437.5J

PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then

PE = -mgh*sin25

PE = -16*9.81*1.15sin25

PE = -82.9J

The net work would then be given by

W_{net} = W+PE

W_{net} = 3437.5J-82.9J

W_{net} = 3354.6J

Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J

You might be interested in
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

8 0
3 years ago
1. Which statement about subatomic particles is not true?
igomit [66]

1. Protons and neutrons have the same charge.

Protons have positive charge, equal to e=+1.6\cdot 10^{-19} C, while neutrons have zero charge.

2. mass number

The mass number of an atom is equal to the sum of protons and neutrons inside its nucleus.

3. Atoms are made up of smaller particles.

According to Dalton's theory, atoms are the smallest particles that make matter, and they are indivisible and indestructible, so they are NOT made up of smaller particles.

4. a solid sphere

In Dalton's theory, atoms are not made of smaller particles, so we can think them as solid spheres.

5. J. J. Thomson

In his experiment with cathode ray tubes, JJ Thomson demonstrated the existance of the electrons, which are negatively charged particles inside the atom. In his model of the atom (plum-pudding model), Thomson thought the atom consists of a uniform positive charge and the electrons are located inside this positive charge.

6. An electron has the same amount of energy in all orbitals.

In fact, each orbital corresponds to a different energy level: the farther the orbital from the nucleus, the higher the energy of the electrons contained in that orbital.

7. A hydrogen atom in heavy water has an extra neutron.

Heavy water is a type of water that contains deuterium, which is an isotope of the hydrogen consisting of one proton and one neutron (so, one extra neutron).

8. The glowing beam was always deflected by charged plates

In his cathode's ray tube experiment, Thomson shows that the beam of unknown particles (= the electrons) were deflected by charge plates, so the particles had to be also electrically charged.

9. electrons move to a lower energy level

When electrons move from a higher energy level to a lower energy, they emit a photon (light) of energy equal to the difference in energy between the two energy levels.

10. orbital

In quantum mechanics, electrons in the atom are not precisely located, since we cannot determine their exact position and velocity at the same time. Therefore, we can only describe regions of space where the electrons have a certain probability to be found, and these regions of space are called orbitals.

11. 14

According to Dalton's theory, the proportions of the reactants must be respected in order to form the same compound. Therefore, we can write:

2 g: 4 g = X : 28 g\\X=\frac{2 g \cdot 28 g}{4 g}=14 g

12. negative charge, found outside the nucleus

Electrons are particles with negative charge of magnitude e=-1.6\cdot 10^{-19}C that orbit around the nucleus. The nucleus, instead, consists of protons (positively charged, with charge opposite to the electron) and neutrons (neutrally charged).

13. move from higher to lower energy levels

When electrons move from a higher energy level to a lower energy inside a neon atom, they emit a photon (which is light) whose energy is equal to the difference in energy between the two energy levels.

14. atomic number from its mass number

In fact:

- the atomic number of an atom (Z) is equal to the number of protons inside the nucleus

- the mass number of an atom (A) is equal to the sum of protons+neutrons inside the nucleus

Therefore, we can find the number of neutrons in the nucleus by calculating the difference between A and Z:

Number of neutrons = A - Z

15. None of them

None of these examples is a good analogy to describe the location of an electron in an atomic orbital: in fact, the position of an electron in an orbital cannot be precisely described, we can only describe the probability to find the electron in a certain position, and none of these example is an analogy of this model.

8 0
3 years ago
You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul
photoshop1234 [79]
1) First of all, let's find the resistance of the wire by using Ohm's law:
V=IR
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega

2) Now that we have the value of the resistance, we can find the resistivity of the wire \rho by using the following relationship:
\rho =  \frac{RA}{L}
Where A is the cross-sectional area of the wire and L its length.
We already have its length L=2.90 m, while we need to calculate the area A starting from the radius:
A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2

And now we can find the resistivity:
\rho =  \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}=  1.48 \cdot 10^{-7}\Omega \cdot m
7 0
3 years ago
You need to measure the lengthof you driveway.which measuring tool would produce accurate results that would expect to be the sa
Aleonysh [2.5K]

C a meter stick with only centimeters

D a ruler with millimeters and centimeters

D would be to the nearest half milimeter. Take some time to measure with a 2 inch ruler. Would you really need to know the length to half a mil ?

3 0
3 years ago
Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges
poizon [28]

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

7 0
2 years ago
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