10 x 4^2 = 160 / 8..

V = 20m/s...

...x 8 = 100 miles,meters, metric what ever m stands for after 8 seconds.

This is my guess since the problem says 4m/s^2

V= distance/ ST (traveled/used)

8 0

2 years ago

What is the temperature of a star (in Kelvin) if its peak wavelength is 150 nm (that is, 150 x 10^-9 m)?

liq [111]

Answer:

19320 K

Explanation:

The temperature of a star is related to its peak wavelength by Wien's displacement law:

$T=\frac{b}{\lambda}$

where

T is the absolute temperature at the star's surface

$b=2.898 \cdot 10^{-3}m\cdot K$ is Wien's displacement constant

$\lambda$ is the peak wavelength

Here we have

$\lambda=150 nm = 150\cdot 10^{-9}nm$

Substituting into the equation, we find

$T=\frac{2.898\cdot 10^{-3}}{150\cdot 10^{-9}}=19320 K$

7 0

3 years ago

Suppose that a simple pendulum consists of a small 81 g bob at the end of a cord of negligible mass. If the angle θ between the

MariettaO [177]

Answer:

(a) 0.115 m

(b) 2.08 x 10^-5 J

Explanation:

mass of bob, m = 81 g = 0.081 kg

The equation of oscillation is given by

θ = 0.068 Cos {9.2 t + Ф}

Now by comparison

The angular velocity

ω = 9.2 rad/s

(a) $\omega^{2} =\frac{g}{L}$

where, L be the length of the pendulum

$L =\frac{g}{\omega ^{2}}$

$L =\frac{9.8}{9.2 \times 9.2}$

L = 0.115 m

(b) A = L Sinθ

A = 0.115 x Sin 0.068

A = 7.8 x 10^-3 m

Maximum kinetic energy

K = 0.5 x mω²A²

K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6

K = 2.08 x 10^-5 J

8 0

2 years ago