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enyata [817]
3 years ago
5

A revolutionary war cannon, with a mass of 2240 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 131 m/s aft

er it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?Answer in units of m/s.
Physics
1 answer:
My name is Ann [436]3 years ago
8 0

Answer:

v' = -0.0906 m/s

Explanation:

given,

mass of cannon, M = 2240 Kg

mass of the ball, m = 15.5 Kg

speed of cannon ball, v = 131 m/s

speed of he cannon = ?

initial speed of cannon and the cannon ball is equal to 0 m/s

using conservation of energy

(M+m)V = M v' + m v

(M+m) x 0= 22400 v' + 15.5 x 131

22400 v' = -2030.5

v' = -0.0906 m/s

negative sign represent the canon will move in opposite direction of the ball.

hence, speed of cannon is equal to 0.0906 m/s

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omeli [17]

The word displacement implies that an object has moved, or has been displaced. Displacement is defined to be the change in position of an object.

Displacement is defined as the act of moving someone or something from one position to another or the measurement of the volume replaced by something else

8 0
3 years ago
41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo
Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

\phi = BAcos \theta

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0
3 years ago
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail
Alik [6]

Answer:

1.34\cdot 10^{-16} C

Explanation:

The strength of the electric field produced by a charge Q is given by

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

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C

4 0
3 years ago
amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i
Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,                                  

weight of equipment = 6 kN      

coefficient of kinetic friction = 0.05          

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²                    

Work done by the camper = ?                

actual acceleration acting a'      

m a = m a' - μ mg            

a' = a + μ g                      

a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

Work done = Force x distance

F = m a'                                                            

F = \dfrac{6000}{9.8} \times 0.69

F = 422.44897 N                            

Work done = F x d                          

Work done = 422.44897 x 1000

Work done = 422449 J                  

Work done = 422.45 kJ            

4 0
3 years ago
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