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2 years ago

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Tahj walks 28 m south from Imhotep's front doors in order to get into his mothers car. If it takes him 7 s, what was average vel

ocity?

2 answers:

DochEvi [55]2 years ago

7 0

The correct answer is 4

In-s [12.5K]2 years ago

5 0

Using a gas power heater to warm a room

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A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Tresset [83]

Answer:

$\theta_1 = 0.400^o$

$\theta_2 =0.378^o$

Explanation:

From the question we are told that

The number of slits per cm is k = $161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m$

The order of the maxima is n = 1

The wavelength are $\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m$

The spacing between the slit is mathematically represented as

$d = \frac{ 0.01}{k}$

=> $d = \frac{ 0.01}{161}$

=> $d = 6.211 *10^{-5} \ m$

Generally the condition for constructive interference is

$n\lambda = d \ sin \theta$

At $\lambda_1$

$\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_1 = 0.400^o$

At $\lambda_2$

$\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_2 =0.378^o$

4 0

2 years ago

What is the volume of a 72 g of copper if copper has a density of 9 g/cm3

azamat

Answer:

8 centimetre cube

Explanation:

because volume is equal to mass per densiry

6 0

3 years ago

A. Community<br> B. Organism<br> C. Population <br> D. Individual

dimulka [17.4K]

Answer:

i think its B or D

Explanation:

its Individual or Organism

3 0

2 years ago

Read 2 more answers

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

vichka [17]

The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

3 0

2 years ago

Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of 2

Sidana [21]

Answer:

The work done on the wagon is 37 joules.

Explanation:

Given that,

The force applied by Charlie to the right, F = 37.2 N

The force applied by Sara to the left, F' = 22.4 N

We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :

$F_n=F-F'$

$F_n=37.2-22.4$

$F_n=14.8\ N$

Work done on the wagon is given by the product of net force and displacement. It is given by :

$W=F_n\times d$

$W=14.8\ N\times 2.5\ m$

W = 37 Joules

So, the work done on the wagon is 37 joules. Hence, this is the required solution.

7 0

2 years ago

Read 2 more answers