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3 years ago

9

A physics book is thrown horizontally at a velocity of 5.0 m/s from the top of a cliff 78.4 m high. How long does the book take

to reach the bottom of the cliff .

1 answer:

xenn [34]3 years ago

3 0

The book's vertical position in the air is

$y=78.4\,\mathrm m-\dfrac12gt^2$

where $g=9.80\,\frac{\mathrm m}{\mathrm s}$ . It reaches the ground when $y=0$ , at a time $t$ such that

$0=78.4\,\mathrm m-\dfrac12gt^2\implies t=4.00\,\mathrm s$

So it takes the book 4 seconds to reach the bottom. The given initial velocity is irrelevant since it only has a horizontal component; vertically, the book is starting from rest.

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A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

Vesnalui [34]

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere= $\rho$

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

$Density=\frac{mass}{volume}$

$Mass=Density\times volume=Constant$

Therefore, $\rho_1 V_1=\rho_2V_2$

Volume of sphere= $\frac{4}{3}\pi r^3$

Using the formula

$\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3$

$\rho R^3=\rho_2\times \frac{R^3}{8}$

$\rho_2=8\rho$

Hence, the density of the compressed sphere= $8\rho$

Option a is correct.

7 0

2 years ago

Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.

iren [92.7K]

Answer:

F_n = 5.65E-11 N

d = 1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

$F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}$

For $F_n$

$F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}$

Dividing the above equations we get

$\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N$

F_n = 5.65E-11 N

$F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m$

d = 1.20682E-31 m

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Arada [10]

Answer:

A hypothesis is a basically a theory proposed to a subject or refrence to an act with limited evidences.

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