If τ=r×F then F.τ is equal

A common flashlight bulb is rated at 0.23 a and 2.9 v (the values of the current and voltage under operating conditions). if the

allochka39001 [22]

The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius

6 0

3 years ago

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think

dangina [55]

Answer:

4.44 rpm

Explanation:

$\omega$ = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = $\frac{3138000}{2}\ m$

R = Radius of arm = 6 m

The acceleration due to gravity is given by

$g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2$

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

$a_c=\omega^2R$

$a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s$

Converting to rpm

$1\ rad/s=\frac{60}{2\pi}\ rpm$

$0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm$

The angular speed of the arm is 4.44 rpm

8 0

3 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

4 years ago

The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from

Scorpion4ik [409]

$V= \frac{S}{t}$
$t= \frac{S}{V}$ <u />
$t= \frac{S}{c}$
$t= \frac{6.1*10^{12}}{299792458}$
$t=20347.4098071s$

It takes 20347.4098071s for light from the sun to reach Pluto.
The 6.1*10^9 is replaced by 6.1*10^12 on line 4 because we convert the distance from km to m.
c = speed of light. If a different value was given in the previous question then use that instead of the value I used to do the final calculation.

3 0

4 years ago

What does weight require?

kherson [118]

Answer:

kg

Explanation:

easy question to be honest

7 0

3 years ago