Calculate the weight of a 25kg object?

A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of fr

STatiana [176]

Answer:62.11 s

Explanation:

Given

weight of man =730 N

mass of man $m_1=\frac{730}{9.8}=74.48 kg$

radius of pound r=5 m

mass of book $m_2=1.2 kg$

velocity of book $v_2=5 m/s$

let $v_1$ be the velocity of man

conserving momentum

$0=m_1v_1+m_2v_2$

$0=74.48\times v_1+1.2\times 5$

$v_1=\frac{6}{74.48}=0.0805 m/s$

time taken by man to reach south end

$t=\frac{5}{0.0805}=62.11 s$

5 0

4 years ago

Q: Compare and contrast the different parts of the Electromagnetic Spectrum. Evaluate the usefulness of each part and decide whi

LUCKY_DIMON [66]

The infrared and ultraviolet parts of the electromagnetic spectrum can be used in the field of chemical analysis.

The sections of the EM spectrum are named in order of increasing energy gamma rays X-rays ultraviolet-visible light infrared, and radio waves. Microwaves used in microwave ovens are a subsection of the radio segment of the EM spectrum.

Electromagnetic energy travels in waves and covers a wide spectrum from very long radio waves to very short gamma rays. The human eye can only see a small portion of this spectrum known as visible light. The electromagnetic spectrum includes all frequencies of electromagnetic radiation that carry energy and travel through space in the form of waves. Low frequencies and long wavelengths make up the radio spectrum.

Learn more about The Electromagnetic spectrum here:-brainly.com/question/23423065

#SPJ1

3 0

1 year ago

A parallel RLC resonant circuit has a resistance of 200 Ω. If it is known that the bandwidth is 80 rad/s and the lower half-powe

d1i1m1o1n [39]

Answer:

L= 3.6mH

C =9.9 microfarad

Explanation:

Resonant frequency fr

fr = fl + 1/2 BW

fr = 800+ 1/2×80

=800+40

=840 rad/s

Bandwidth (BW)

BW = fr/Q

Q = quality factor

Q= fr/ BW

Q = 840/80

Q= 10.5

Quality factor = R/Xl

Xl = inductive reactance

Xl = R/Q

Xl = 200/10.5

Xl = 19.05 ohms

Xl =2πfL

L= Inductance

L = Xl /2πf

L =19.05/5278.56

L= 3.6mH

Capacitor C

1= fr^2 × 4π^2LC

C = 1/fr^24π^2L

C = 1/100307.5

C= 9.9microfarad

5 0

3 years ago

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the

Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) $8.40A/m^2$

(c) $\rho =15.52\times 10^{-9}ohm-m$

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

$A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2$

Resistance $R=11.9mohm=11.9\times 10^{-3}ohm$

Potential difference V = 33.7 volt

(A) current is equal to

$i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A$

(B) Current density is equal to

$J=\frac{i}{A}$

$J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2$

(c) Resistance is equal to

$R=\frac{\rho l}{A}$

$11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}$

$\rho =15.52\times 10^{-9}ohm-m$

4 0

3 years ago

1. Which of the following is correct about the sampling distribution of the sample mean

joja [24]

Answer:

In my opinion I think that the answer is C sorry If I get this wrong.

5 0

3 years ago