An object is placed 12.5 cm from a lens of focal length 22.0 cm. What is the image distance?

a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

Katyanochek1 [597]

Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
ANSWER: 3 m/s²

Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration

Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

ANSWER: 240 N 

Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²

Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m

ANSWER: d = 13.5 m

3 0

3 years ago

Technician A says that other temperature sensors that operate like the ECT include transmission fluid temperature (TFT), and eng

PIT_PIT [208]

Answer Choices:

a. Technician A only

b. Technician B only

c. Both technicians A and B

d. Neither technician A nor B

Explanation:

a. Technician A only

5 0

3 years ago

light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur

VashaNatasha [74]

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

$t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}$

where;

n is refractive index of soap film = 1.33

$t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t = 225.56 *10^{-9} \ m\\\\t = 225.56 \ nm$

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

4 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

3 years ago

State the laws of vibration of a stringed Instrument

olchik [2.2K]

If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.

7 0

3 years ago