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3 years ago

11

The mass of a lift is 600kg.the maximum tensile force that the cable supporting the lift can withstand is 7kN.Calculate:maximum

upwards acceleration allowed?

1 answer:

Elena-2011 [213]3 years ago

7 0

Use F=ma. F is the tension in the cable 7e3N. m is 600kg. Solve for a:
a=7e3/600=11.67m/s^2

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A uranium nucleus 238U may stay in one piece for billions of years, but sooner or later it decays into an α particle of mass 6.6

ale4655 [162]

Answer: The recoil speed is - 8.9604. $10^{4}$ m/s.

Explanation: According to the Third Law of Newton, every action has an oppsite and equal reaction, and the Second Law of Newton, Force=mass·acceleration. Acceleration is a variation in velocity by any given time, so Force = mass·velocity·time.

Combining the two laws, there is : m1·v1 = - m2·v2. This is the Law of Conservation of Momentum.

Substituting and calculating:

v2 = - ( $\frac{m1}{m2}$ ) · v1

v2 = - $\frac{6.64.10^{-27} }{3.88.10^{-25} }$ · 5.24. $10^{6}$

v2 = - 8.9604. $10^{4}$

The recoil speed of the thorium nucleus is - 8.9604. $10^{4}$ m/s.

8 0

4 years ago

A series RLC circuit is driven by a 1.0-kHz oscillator. The circuit parameters are Vms = 12 V, L = 5.0 mH, C= 4.0 uF, and R=102.

alexgriva [62]

Answer:

C) $12.3$ volts

Explanation:

f = frequency of oscillator = 1 kHz = 1000 Hz

$V_{rms}$ = 12 Volts

L = Inductance of Inductor = 5 mH = 0.005 H

$X_{L}$ = Inductive reactance

Inductive reactance is given as

$X_{L}$ = $2\pi fL$

$X_{L}$ = $2(3.14) (1000)(0.005)$

$X_{L} = 31.4$

$X_{C}$ = Capacitive reactance

Capacitive reactance is given as

$X_{C}=\frac{1}{2\pi fC}$

$X_{C}=\frac{1}{2(3.14)(1000)(4\times 10^{-6})}$

$X_{C} = 39.8$

Impedance of the circuit is given as

$z = \sqrt{R^{2}+(X_{L} - X_{C})^{2}}$

$z = \sqrt{102^{2}+(31.4 - 39.8)^{2}}$

$z = 102.35$

Rms Current flowing is given as

$i = \frac{V_{rms}}{z}$

$i = \frac{12}{102.35}$

$i = 0.12$ A

Rms potential difference across the resistor is given as

$V = i R$

$V = (0.12) (102)$

$V = 12.3$ volts

5 0

3 years ago

A tank, shaped like a cone has height 6 meters and base radius 5 meters long. It is placed so that the circular part is upward.

AlekseyPX

The work required to pump out the liquid is 2.35 MJ.

What is work?

Work can be defined as the scalar product of the force and the displacement.

The formula to calculate the work done W is,

$W=F.d$

where F is the force and d is the displacement.

Note: It is assumed that the weight of one cubic unit of the liquid is $10000 \text{ N/m}^3$ .

Break down the vertical tank into parallel disc elements whose thickness is dh and are located at a height h from the ground. The radius of the element at height h is r.

It is given that the height of the tank is 6 m and the base radius is 5 meters. Using these, the ratio of the radius r at height h to the height h by using the triangle similarity criterion is,

$\frac{h}{r}=\frac{6}{5} \\ r=\frac{5h}{6}$

Using the above value of r, the volume of the element is given by,

$V= \pi r^2dh\\V=\pi (\frac{5h}{6})^2dh\\V= \frac{25\pi h^2}{36} dh$

The weight of the volume element is the force required.

The weight of one cubic unit of the liquid is $10000 \text{ N/m}^3$ .

So the weight F of the volume element at height h is given by,

$F=10000 (\frac{25 \pi h^2}{ 36})dh\\F= 6944.44\pi h^2dh$

At the height h, the difference in the height of the top of the tank and the volume element is (6-h). So the work required to pump out the volume of liquid at height h is given by the integrating the equation,

$W=6944.44\pi\int\limits^6_0 {(6h^2-h^3)} \, dx \\\\W=69.44\pi \left[ 2h^3-\frac{h^4}{4}\right]^6_0\\W=2,.35\times10^6\text{ J}$

The value of the W obtained is 2.35*10^6 J or 2.35 MJ.

Learn more about work here:

brainly.com/question/16627899

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8 0

2 years ago

an electric soldering iron takes 5 amperes at 110 volts.what is the power used in watts?in kilowatts?

vichka [17]

Answer:

550 W, 0.55 kW

Explanation:

Power = Current * Voltage

P = 5A * 110V

P = 550 W

There's 1000 W in 1 kW, so P = 0.55 kW.

7 0

3 years ago

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

8 0

4 years ago