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ozzi
3 years ago
13

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

the bullet rises to a height of 10 cm along a circular arc with a 18 cm radius.
Assume: The entire track is frictionless. A bullet with a m1 = 30 g mass is fired horizontally into a block of wood with m2 =
4.8 kg mass.
The acceleration of gravity is 9.8 m/s^2.
Calculate the total energy of the composite system at any time after the collision. Answer in units of J.
Taking the same parameter values as those in Part 1, determine the initial velocity of the bullet. Answer in units of m/s.
Physics
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

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almond37 [142]

a. 11.28\Omega

The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:

R_{eq}=R_1 + R_2

In this circuit, we have

R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega

Therefore, the equivalent resistance is

R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega

b. 5.8 V, 3.2 V

First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:

I=\frac{V}{R_{eq}}

where V = 9.00 V and R_{eq}=11.28 \Omega. Substituting,

I=\frac{9.00 V}{11.28 \Omega}=0.8 A

Now we can calculate the potential difference across each resistor by using Ohm's law again:

V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V

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4 0
3 years ago
A rock exerts 5000 Pa of pressure on the ground. If the rock weighs 250 N, how much area is in contact with the ground?
ololo11 [35]

P = F/S - S = F/P = 250/5000= 0.05 m2

4 0
3 years ago
Read 2 more answers
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo
Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

W_{work}=2.67*10^{-3}J

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

W_{work}=F_{force}*D_{distance}Cos\alpha  \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J

8 0
3 years ago
Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp
hammer [34]

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = \frac{dh}{dt}=22\ cm/s

\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3

Differentiating with respect to time

\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s

∴ Rate is the sand leaving the bin at that​ instant is 159241.048 cm³/s

5 0
3 years ago
A string that is restricted at both ends has a length of 1.50 m. what is the wavelength of the string’s fundamental frequency?
lara31 [8.8K]
Since it is restricted at both ends, λ/2 = length of string

λ/2 = 1.5m
λ = 1.5*2 = 3m
6 0
3 years ago
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