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Andrei [34K]
2 years ago
13

Sir Marvin decided to improve the destructive power of his cannon by increasing the size of his cannonballs. Sir Seymour kept hi

s cannonballs the same size, but improved his powder to provide more velocity. a) Which knight will have the more destructive cannon? Why?
Physics
1 answer:
maria [59]2 years ago
5 0
We really can't tell from the given information. 
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.

If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more. 

I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.

                K.E.  =  (1/2) (mass) (speed-SQUARED) .

We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.

But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.

That's 4 times as much destructive capability as before.  

So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
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A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
2 years ago
A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

7 0
2 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th
mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = \frac{\textup{1}}{\textup{2}}  × C × V²

on substituting the respective values, we get

U = \frac{\textup{1}}{\textup{2}}  × 1.27 × ( 59.9 × 10³ )²

or

U = 1794.005 × 10⁶ J

7 0
2 years ago
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