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grandymaker [24]
3 years ago
12

A steel wire of cross_ sectional area 2x10-5 m2 is stretched through 2mm by a force of 4000 N. find the young's modulus of the w

ire . the length of the wire is 2m.
please answer me.​
Physics
1 answer:
UNO [17]3 years ago
7 0

Answer:

hi how are you doing today Jasmine

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One student bangs two bricks together
Romashka-Z-Leto [24]

Answer:

This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time

Explanation:

I hope this is what you need

PLEASE MAKE ME BRAINLIEST

7 0
3 years ago
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p
Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁  = (A₂ / A₁)v₂

v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂

substitute v₁  into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

8 0
3 years ago
How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?
Luda [366]

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375  x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375  x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

4 0
3 years ago
When a rubber band is pulled back on your finger but not yet let go how is that potential energy?
loris [4]
It is potential energy because the band is not in movement, th band has the potential to move.
4 0
3 years ago
A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite
sladkih [1.3K]

Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

7 0
3 years ago
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