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Llana [10]
3 years ago
14

A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.

0 g and a speed of 310 m/s
A) How high will the block rise into the air after the bullet becomes embedded in it?
B) What is the initial momentum of the system?
C) What is the speed of the block just after the bullet becomes embedded in it?
Physics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s

Momentum = 310*0.022

Momentum = 6.82 Ns.

If the bullet gets embedded they will both have common velocity v

6.82 = (0.022+1.40)v

6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

Where tita is 90°

H = (4.796² * sin²(90))/(2*9.81)

H =( 23.001616*1)/19.62

H = 1.172 metres

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Explanation:

The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".

There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.

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In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a
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Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

centripetal acceleration (a_c)=9\times 10^22 m/s^2

we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

v=21.84\times 10^5 m/s

(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

a_c=10^4\times 9\times 10^{22} m/s^2

a_c=9\times 10^{26} m/s^2

v=\sqrt{r\times a_c}

v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}

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3 years ago
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
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Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

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17.Explain the different ways that an object can become electrically charged.
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17.

There are three different methods for charging objects:

- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.

- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.

- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.

18.

Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)

E=k\frac{q}{r^2}

where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.

Also, the direction of the field is determined by the sign of the charge:

- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)

- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)

19.

Electrical force is given by:

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the two charges, and r their separation.

Gravitational force is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.

Similarities between the two forces:

- Both are inversely proportional to the square of the distance between the two objects, r

- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)

- Both forces have infinite range

Differencies between the two forces:

- The electric force can be either attractive or repulsive, while the gravitational force is attractive only

- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G

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