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andrezito [222]
3 years ago
11

How much tension must a rope withstand if it is used to accelerate a 1210- kg car horizontally along a frictionless surface at 1

.20 m/s squared?
Physics
1 answer:
Ratling [72]3 years ago
4 0
Newton's second law states that the force F applied to the car is the product between the mass of the car m and its acceleration a:
F=ma
But the force applied to the car is the tension of the rope, T, so we have:
T=ma

And so, using the data of the problem we calculate the value of the tension:
T=ma=(1210 kg)(1.20 m/s^2)=1452 N
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sasho [114]

Answer:

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b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m

Explanation:

(a) Radiowave wavelength= λ = c/f

As we know, Radiowave speed in the air = c = 3 x 10^8 m/s

f = frequency = 604 kHz = 604 x 10^3 Hz

Hence, wavelength = (3x10^8/604x10^3) m

λ = 496.69 m

So the height of the antenna BROADCASTING AT 604 kHz =  λ /4 = (496.69/4) m

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(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz  (1kHZ = 1000 Hz)

Hence, wavelength =  λ = (3 x 10^8/1710 x 10^3) m

 λ= 175.44 m

So, height of the antenna =  λ /4 = (175.44/4) m

= 43.86 m  

5 0
4 years ago
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Explanation:

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E = 401.3 eV

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