What does more damage; a slow semi truck, or a fast sports car

A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra

Sonja [21]

Answer:

Intensity, $I=1.101\ W/m^2$

Explanation:

Power of the light bulb, P = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

$I=\dfrac{P}{A}$

$I=\dfrac{P}{4\pi r^2}$

$I=\dfrac{40\ W}{4\pi (1.7\ m)^2}$

$I=1.101\ W/m^2$

So, the intensity of light is $1.101\ W/m^2$ .

6 0

4 years ago

HELP ASAP please...

luda_lava [24]

The answer to the question is A

7 0

3 years ago

You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft.

Nookie1986 [14]

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

$V_{slip$ = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

$a_c$ = 2( $V_{slip$ × ω )

we substitute

$a_c$ = 2( 8.8002 ft/s × 2.5133 rad/s )

$a_c$ = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

7 0

3 years ago

A bat hits a moving baseball. If the bat delivers a net eastward impulse of 0.80 N-s and the ball starts with an initial horizon

AURORKA [14]

Answer:

0.084 kg

Explanation:

I = 0.80 N-s (East wards) = 0.80 i N-s

u = 3.8 m/s = - 3.8 i m/s

v = 5.7 m/s = 5.7 i m/s

Let m be the mass of bat.

I = m (v - u)

0.8 i = m ( 5.7 i + 3.8 i)

0.8 i = m x 9.5 i

m = 0.084 kg

8 0

3 years ago

You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6

marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

$L= \frac{\mu_o N^2 A}{l}$

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

$L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H$

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

$E = \frac{1}{2} LI^2\\\\$

Where;

I is the current in the solenoid

$E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A$

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

$U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3$

3 0

3 years ago