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3 years ago

6

11 kg is a familiar weight for a bag of flour. You are baking cookies for a Save The Rain Forest fund drive. It takes 500 g of f

lour to make one batch of cookies. How many batches of cookies can you make with one bag of flour

1 answer:

arlik [135]3 years ago

8 0

Answer: 22 batches.

Explanation:

Given that 11 kg is a familiar weight for a bag of flour. Also, it is given that It takes 500 g of flour to make one batch of cookies.

How many batches of cookies can you make with one bag of flour

Let's first convert 11 kg into grams (g) by multiplying it by 1000

11 × 1000 = 11000 g

Divide 11000 by 500

11000/500 = 22

Therefore, 22 batches of cookies can be made with one bag of flour.

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During the class prize-giving ceremony, Anand clapped his hands hard while Kumar clapped his hands softly. Everybody could hear

algol13

Answer:

C - higher volume

Explanation:

The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .

But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .

3 0

3 years ago

AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0Ã108m/s. What are the w

Westkost [7]

Answer:

The wavelength of these signals is as follow:

Wavelength of 550 kHz is 545.45 m
Wavelength of 1600 kHz is 187.5 m

Explanation:

Given that:

Frequency = 550 kHz & 1600 kHz

Velocity = 3.0 x 10⁸ m/s

As we know that frequency is expressed by the following equation:

Frequency = Velocity / Wavelength ---- (1)

For 550 kHz:

The equation can be rearranged as

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (550 x 1000 Hz)

Wavelength = 545.45 m

For 1600 kHz:

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (1600 x 1000 Hz)

Wavelength = 187.5 m

5 0

3 years ago

A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a

Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

$V_c = l^3$

$V_c = 6^3 = 216cm^3$

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

$V_d = \pi r^2*l$

$V_d = \pi (\frac{2}{2})^2*6$

$V_d = 6\pi cm^3$

In this way the net volume would be

$\Delta V = V_c-V_d$

$\Delta V = 216cm^3-6\pi cm^3$

$\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3$

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

$F= mg$

$m = \frac{F}{g}$

$m = \frac{6.6}{9.8}$

$m = 0.673kg$

PART A) From the relation of density as a unit of mass and volume we have to

$\rho = \frac{m}{V}$

$\rho = \frac{0.673}{197.15*10^{-6}}$

$\rho = 3413.64kg/m^3$

PART B) To find the weight of the cube then we apply the ratio of

$W = mg$

$W = V\rho g$

$W = (216*10^{-6})(3413.64)(9.8)$

$W = 7.22N$

3 0

3 years ago

A fish finder uses a sonar device that sends 20,000-Hz sound pulses downward from the bottom of the boat, and then detects echoe

mrs_skeptik [129]

Answer: The minimum time between pulses (in fresh water)

= 0.3106 s

Explanation:

To calculate the speed of echoes sounds, we will use

Speed = 2x/t

Where x = distance and t = total time.

Please note that:

sound travels at 343 m/s in air. But it travels at 1,481 m/s in water (almost 4.3 times as fast as in air);

Total distance = 2 × 230 = 460m

Using speed of sound in water

1481 = 460/t

t = 460/1481

t = 0.3106

3 0

2 years ago

In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sec

Mrac [35]

Answer:

$F_s=1075.9493\ N$

Explanation:

Given:

area of piston on the smaller side of hydraulic lift, $a_s=0.075\ m^2$

area of piston on the larger side of hydraulic lift, $a_l=0.237\ m^2$

Weight of the engine on the larger side, $W_l=3400\ N$

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

$P_s=P_l$

$\frac{F_s}{a_s}=\frac{W_l}{a_l}$

$\frac{F_s}{0.075} =\frac{3400}{0.237}$

$F_s=1075.9493\ N$ is the required effort force.

5 0

3 years ago

Read 2 more answers