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2 years ago

5

When a cannon fires a cannonball, the cannon will recoil backward because the:

1 answer:

gulaghasi [49]2 years ago

4 0

C) total linear momentum of the ball and cannon is conserved.

Basically it happens that in the beginning before there is a momentum acting on the two bodies, these are a unique system. Here the total momentum of the System is 0. However, when the positive momentum of the cannonball is added, the system will be immediately affected by a negative momentum which will pull back the cannon. Could this be extrapolated as a condition of Newton's third law.

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A plucked violin string carries a traveling wave given by the equation f(x,t)=asin[b(x−ct)+ϕi], with a = 0.00580 m , b = 33.05 m

Viktor [21]

Answer:

A) Φ = 0 , B) T = 7.76 s

Explanation:

A) to find the value of the phase constant replace the value

0 = a sin (b (0- 0) + Φ)

0 = sin Φ

Φ = sin⁻¹ 0

Φ = 0

B) the period is defined by time or when the movement begins to repeat itself

So that the sine function is repeated when the angle passes 2pi

b (x- ct) = 2pi

If we are at a fixed point x = 0

b c t = 2pi

t = 2π / bc

Let's calculate

T = 2π / (33.05 245)

T = 7.76 s

0 0

2 years ago

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What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the <u>electric constant</u>

$q=1.6(10)^{-19}C$ is the <u>electric charge of the hydrogen nucleus</u>, which is positive

$r=5.292(10)^{-11}m$ is the <u>distance</u>

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

5 0

3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

Which one?? please someone quick!

LiRa [457]

Answer:

i think its second law of motion.

Explanation:

4 0

2 years ago

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A train’s mass is 18181.81 kg. What is its weight?

wel

Answer:

The trains mass in pounds would be 40084.029 if you would round it to the hundreths

Explanation:

7 0

2 years ago

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