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4 years ago

HELPPPPP ASAPPP!!!!!!!

Physics

1 answer:

bezimeni [28]4 years ago

4 0

Answer:

The tension T in each of the cables, T = 98 N

Explanation:

Given,

The mass of the loudspeaker, m = 20.0 Kg

The suspended distance of the loudspeaker, d = 2.20 m

The length of each cable, l = 2.9 m

The tension experienced by the cable is given by the formula,

T = mg + ma newton

Where,

g - acceleration due to gravity

a-acceleration produced by the stretched cable

Since the stretched cable and loudspeaker is not moving, then the tension force acting on the cable is equal to the force of gravity.

T = mg newton

Substituting the values in the above equation

T = 20

= 20.0 Kg x 9.8 m/s²

= 196 N

This force is equally supported by the two cables, then the tension in each cable is

T = 196 N / 2

= 98 N

Hence, the tension T in each of the cables, T = 98 N

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alexandr1967 [171]

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3 years ago

A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕

Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0

3 years ago

An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t

Vera_Pavlovna [14]

Answer:

KE₂ = 4KE₁

Explanation:

KE₁ = ½mv₁²

KE₂ = ½mv₂²

KE₂ = ½m(2v₁)²

KE₂ = 4(½mv₁²)

KE₂ = 4KE₁

7 0

3 years ago

A plane electromagnetic wave, with wavelength 5 m, travels in vacuum in the positive x direction with its electric vector E, of

Dafna1 [17]

Answer:

Explanation:

E₀ = 229.1 V/m

E = E₀ / √2 = 229.1 / 1.414 = 162 V/m

B = E / c ( c is velocity of em waves )

= 162 / (3 x 10⁸) = 54 x 10⁻⁸ T

rate of energy flow = ( E x B ) / μ₀

= 162 x 54 x 10⁻⁸ / 4π x 10⁻⁷

= 69.65 W per m².

5 0

3 years ago

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Mademuasel [1]

Here it is. *WARNING* VERY LONG ANSWER

________________________________________...
11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) 

The change in PE =mgh=5*9.8*12=588 J 
______________________________________... 
12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. 

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, 

Their KE as they crossed the line=(1/2)Mv^2 

Their KE as they crossed the line=0.5*550*(15.98)^2 

Their KE as they crossed the line is 70224.11 J 

______________________________________... 
13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m 

She trips and drops the spare tire of mass = m = 10.0 kg, 

The tire rolls down the hill with an intial speed = u = 2.00 m/s. 

The height of top of the next hill = h = 5.00 m 

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 

Initial total mechanical energy =mgH+(1/2)mu^2 

Suppose the final speed at the top of second hill is v 

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 

As mechanical energy is conserved, 

Final total mechanical energy =Initial total mechanical energy 

mgh+(1/2)mv^2=mgH+(1/2)mu^2 

v = sq rt [u^2+2g(H-h)] 

v = sq rt [4+2*9.8(20-5)] 

v = sq rt 298 

v =17.2627 m/s 

The speed of the tire at the top of the next hill is 17.2627 m/s 
______________________________________... 
14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. 

a.)The mass of bean = m = 2.0 g 

Height up to which the been jumps = h = 1.0 cm from hand 

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg 

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s 
_____________________________ 
15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. 

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s 
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16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, 

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 
______________________________________... 

EDIT 
1.) A train is accelerating at a rate = a = 2.0 km/hr/s. 

Acceleration 

Initial velocity = u = 20 km/hr, 

Velocity after 30 seconds = v = u + at 

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = 

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr 

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s 
_______________________________- 
2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. 

His acceleration = a =11.1/9=1.233 m/s^2 

Distance he covered = s = (1/2)at^2=49.95 m

7 0

3 years ago