PHYSICS COLLEGE

Which sling can the crane use to lift the 1000kg pipe?A.
800kg rated sling
B. 1000kg rated sling
C. 2000kg rated sling
D. Band C

Answers

Answer 1

Answer:

Answer:

C. 2000kg rated sling

Explanation:

ensures better safety and can carry twice more mass than current mass.

Related Questions

An emf is induced by rotating a 1207 turn, 20.0 cm diameter coil in the Earth's 4.13 10-5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms
If a person’s weight is W on the surface of the earth, calculate what it would be, in terms of W, at the surface of (a) the moon; (b) Mars; (c) Jupiter.
Average wavelength of radio waves
Which is the SI base unit for mass?
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Answers

Answer:

$I=8.48* 10^(-4)\ W/m^2$

Explanation:

Given that,

Frequency of the radio signal, $f=800\ kHz=8* 10^5\ Hz$

It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km

The amplitude of the electric field is, E = 800 mV/m

Let I is the intensity of the radio signal at that point. Mathematically, it is given by :

$I=(E^2_(rms))/(c\mu_o)$

$E_(rms)$ is the rms value of electric field, $E_(rms)=(E)/(√(2) )$

$I=(E^2)/(2c\mu_o)$

$I=((800* 10^(-3))^2)/(2* 3* 10^8* 4\pi * 10^(-7))$

$I=8.48* 10^(-4)\ W/m^2$

So, the intensity of the radio signal at that point is $8.48* 10^(-4)\ W/m^2$ . Hence, this is the required solution.

Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion

Answers

Answer:

It will take 29.31 seconds for the Boxster to catch the Scion

Explanation:

Given the data in the question;

lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A

the distance travelled by car A is

x = $V_(A)$ × t

where $V_(A)$ is the speed of the car and t is time

the distance travelled by car B before reaching car A will be;

x + x₀ = $V_(B)$ × t

Now lets replace x by $V_(A)$ × t

( $V_(A)$ × t) + x₀ = $V_(B)$ × t

x₀ = ( $V_(B)$ × t) - ( $V_(A)$ × t)

x₀ = t ( $V_(B)$ - $V_(A)$ )

t = x₀ / ( $V_(B)$ - $V_(A)$ )

so we substitute

t = 170 m / (24.4 - 18.6)

t = 170 / 5.8

t = 29.31 s

Therefore; it will take 29.31 s for the Boxster to catch the Scion

What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?

Answers

Answer: The average kinetic energy of hydrogen atoms is $1.19562* 10^(-19)J$

Explanation:

To calculate the average kinetic energy of the atom, we use the equation:

$K=(3)/(2)kT$

where,

K = average kinetic energy

k = Boltzmann constant = $1.3807* 10^(-23)J/K$

T = temperature = $5500^oC=[5500+273]K=5773K$

Putting values in above equation, we get:

$K=(3)/(2)* 1.3807* 10^(-23)J/K* 5773K\n\nK=1.19562* 10^(-19)J$

Hence, the average kinetic energy of hydrogen atoms is $1.19562* 10^(-19)J$

A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train