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2 years ago

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A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a densi

ty of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?

1 answer:

cupoosta [38]2 years ago

7 0

Answer:

$N=0.37N$

Explanation:

Mass $m=55g=>0.055kg$

Soapstone Density $\rho_s=3000kg/m^2$

Whisky Density $\rho_w=940kg/m^2$

Generally the equation for Force is mathematically given by

$F=U+N$

Therefore

$N=m*g-(\frac{m}{\rho_s})*\rho_w*g$

$N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}$

$N=0.37N$

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Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed

sergij07 [2.7K]

Answer:

$v_{B0}=15.73 m/s$

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

$p_{0x}=p_{fx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40)$ (1)

y-coordinate

$p_{0y}=p_{fy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40)$ (2)

We can divide equations (2) and (1):

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}$

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)$

$v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)$

$v_{B0}=\frac{1200*25}{1600}*tan(40)$

$v_{B0}=15.73 m/s$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

What a measurement? Write down t its importance inn an our daily life.

balu736 [363]

Answer:

Time, size, distance, speed, direction, weight, volume, temperature, pressure, force, sound, light, energy—these are among the physical properties for which humans have developed accurate measures, without which we could not live our normal daily lives. Measurement permeates every aspect of human life.

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2 years ago

A block of mass 1.5 hangs at the of end of a weight cord suspended from the ceiling.what is the tension in the cord, and with wh

Len [333]

The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

<h3>What is the tension in the cord?</h3>

The tension in the cord is calculated as follows;

T = ma + mg

where;

a is the acceleration of the block
g is acceleration due to gravity
m is mass of the block

T = m(a + g)

T = 1.5(a + 9.8)

T = 1.5a + 14.7

Thus, the tension in the cord is (1.5a + 14.7) N.

If the block is at rest, the tension is 14.7 N.

<h3>Force of the force</h3>

The force with which the cord pulls is equal to the tension in the cord

F = T = m(a + g)

F = (1.5a + 14.7) N

If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.

Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

Learn more about tension here: brainly.com/question/187404

#SPJ1

4 0

6 months ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

or

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago

URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i

STatiana [176]

Answer:

Explanation:

heat lost by water will be used to increase the temperature of ice

heat gained by ice

= mass x specific heat x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186

8372

heat lost = heat gained

1 x 2090 x t = 8372

t = 4°C

There will be a rise of 4 degree in the temperature of ice.

5 0

2 years ago