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ollegr [7]
3 years ago
11

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

that between the large faces (D x L.) 2W. What is the ratio of the resistance as measured between the small faces (WxW) to that between the large faces (DxL)?
Physics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

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platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin
posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

a=g=9.8 m/s^2 is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s

For the diver jumping from 10 m, s = 10 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

s=ut+\frac{1}{2}at^2

And since u = 0, it can be reduced to

s=\frac{1}{2}at^2

For the diver jumping from 5 m, s = 5 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s

For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

5 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
2 years ago
Aiden takes 0.10s to throw a baseball, which leaves his hand with a velocity of 49m/s. The balls acceleration is
Brums [2.3K]
Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
a =  \frac{v}{t}
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
a =  \frac{49}{0.1}  \\ a =  \frac{490}{1}  \\ a = 490
The acceleration is 490 m/s^2
7 0
3 years ago
Which of the following is Not true about the noble gases
Alexandra [31]
Noble gases are not highly reactive
3 0
3 years ago
A student uses the circuit shown to determine the resistance of two identical resistors.
Ivanshal [37]

Answer:

B. 0.552

Explanation:

To find the resistance in the circuit above, u simply divide the current in the circuit by the voltage to get the resistance.

7 0
2 years ago
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