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3 years ago

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A 15.0-kg block is dragged over a rough, horizontal surface by a 75-n force acting at 21° above the horizontal. the block is dis

placed 5.7 m, and the coefficient of kinetic friction is 0.200. (a) find the work done on the block by the 75-n force.

1 answer:

Zepler [3.9K]3 years ago

6 0

Some dogs may inherit a susceptibility to epilepsy.

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o

blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

$P=\rho \times g \times h$

Here

P is the pressure which is given as 68 kPa.
ρ is the density of the oil whose SG is 0.92. It is calculated as

$\rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\$

g is the gravitational constant whose value is 9.8 m/s^2
h is the height which is to be calculated

$P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m$

So the height of column is 7.54m

a-3

By the relation of volume and density

$M=\rho \times V$

Here

ρ is the density of the oil which is 920 kg/m^3
V is the volume of cylinder with diameter 16m calculated as follows

$V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3$

Mass is given as

$M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg$

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

$\Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\$

Now corrected pressure is as

$P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa$

Finding the value of height for this corrected pressure as

$P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m$

The original height of column is 5.98m

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4 years ago

Which of the following is a correct principle of relative-age dating?

elena55 [62]

The anwser choice is A

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4 years ago

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Rita has two small containers, one holding a liquid and one holding a gas. Rita transfers the substances to two larger container

Temka [501]

Answer: liquids take the shape of the container they are in, but have definite volume. like liquids the shape of a gas changes with the container. unlike liquids the volume of a gas changes depending on the container it is in

Explanation:

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4 years ago

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X rays of wavelength 0.0100 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

Fofino [41]

Answer:

a) $=4.84*10^{-12}$

b) $= -2.76*10^{-14} J$

c) $i.e -2.76*10^{-14} J$

d)= 0 and the direction of motion is equal to zero

Explanation:

a) compton shift

$\Delta\lambda = \frac{h}{mc} (1-cos\theta)$

$\Delta\lambda = \frac{6.626*10^{-34}}{9.11*10^{-11}3*10^8} (1-cos180)$

$=4.84*10^{-12}$

b) the new wavelength

$\lambda' = 10.0*10^{-12} +4.84^10^{-12}$

$=14.84*10^{-12} m$

$\Delta E = E' - E$

$=hc[\frac{1}{\lambda'}-\frac{1}{\lambda}]$

$\Delta E = 6.626*10^{-34}*(3*10^8)[\frac{1}{14.84*10^{-12}}-\frac{1}{4.8*10^{-12}}]$

$= -2.76*10^{-14} J$

C)By conservation of energy, the kinetic energy of recoiling electron is equal to the magnitude of energy between the photon energy

$i.e -2.76*10^{-14} J$

d) the angle between the positive direction of motion

$sin\phy = \frac{\lambda_t sin\theta}{\lambda'}$

$=\frac{2.43*10^{-12}sin180}{14.84*10^{-12}}$

= 0

the direction of motion is equal to zero.

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4 years ago

A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/

weqwewe [10]

Answer:Given mass = 4.1kg

Radius = 0.0117m

Velocity V = 8.4m/s

Coefficient of friction = 0.25

Explanation: Below is an attached solution to the problem stated above.

1. The angular acceleration is equal to 524rad/s^2

2. The linear acceleration is equal to 2.45m/s^2

3.the time it takes the ball to begin rolling = 0.98s

4. The distance the ball slides before it begins to roll = 7.05m

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4 years ago