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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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Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
3 years ago
What is the current in a 160V circuit if the resistance is 2Ω?<br> V=<br> I=<br> R=
Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

<u>160</u> = <u>2I</u>

2 2

I = 80A

4 0
2 years ago
A shopper walks westward 5.4 meters and then eastward 7.8 meters
RoseWind [281]

Answer:

13.2 meters

Explanation:

(5.4) + (7.8)

6 0
2 years ago
A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
Any tips on how to get a good grasp of physics and chemistry?
Helga [31]

Answer: Take notes and research into the subject more get extra help like tutoring if needed.

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