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laila [671]
2 years ago
14

A 400 N steel ball is suspended by a light rope fromthe ceiling. The tension in the rope is​

Physics
1 answer:
alina1380 [7]2 years ago
8 0

Answer:

As force = tension in string so tension in string is 400N

Explanation:

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You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
zaharov [31]

Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
2 years ago
The blade of metal cutter is shorter than that scissors of tailors<br><br>​
Naily [24]

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure

8 0
2 years ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
2 years ago
Read 2 more answers
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro
Lerok [7]

Answer:

0.0000045 s

Explanation:

f = Frequency = 8 MHz

Clock cycle is given by

\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s

Time taken for 12 clock cycles

12\times 1.25\times 10^{-7}=0.0000015\ s

Time taken per instruction is 0.0000015 s

In reading and displaying information it requires 3 processes

1 for reading, 1 for searching and 1 for displaying.

3\times 0.0000015=0.0000045\ s

Time taken is 0.0000045 s

6 0
2 years ago
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