All categories

3 years ago

13.4. Young's modulus for iron is 1.9 x 10' Pa. When an iron wire 1.0 m long with a cross-sectional area of 4.0 mm

Physics

1 answer:

Viefleur [7K]3 years ago

6 0

Answer:

0.0129 m

Explanation:

ΔL = FL / (EA)

where ΔL is the deflection,

F is the force,

L is the initial length,

E is Young's modulus,

and A is the cross sectional area.

F = mg = 100 kg × 9.8 m/s² = 9800 N

A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²

ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))

ΔL = 0.0129 m

You might be interested in

3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t

sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

$W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J$

The work done on the puck is 12.3 J.

5 0

4 years ago

___________________ and ___________________________ are two main types of waves.

iris [78.8K]

Answer:

its A

Explanation:

6 0

4 years ago

two forces are acting on a wheelbarrow. One force is pushing to the right and an equal force is pushing to the left. What can yo

ICE Princess25 [194]

B it is not accelerating

3 0

3 years ago

Read 2 more answers

A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rat

Inessa05 [86]

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5

⇒ h = (5/2)*r

⇒ r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π

a) if r = (2/5)*h

⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒ (4/25)(3*h²)(dh/dt) = -105/π

⇒ dh/dt = -875/(4π*h²)

b) if h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒ (5/2)*(3*r²)(dr/dt) = -105/π

⇒ dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒ dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)

⇒ r = 50 cm

⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒ dr/dt = -17.82*10⁻⁴ cm/s

4 0

3 years ago

An astronomer sees two stars in the sky. Both stars are equally far from Earth, but the first star is dimmer than the second sta

kvv77 [185]

Answer: It is smaller than the second star

Explanation: The brightness of a star depends on two factors: The distance from the Earth and size of the star. More the distance from the Earth, the star appears dimmer. And bigger the star, brighter it is.

$b =\frac{L}{4\pi d^2}$

Where L is the luminosity,d is the distance

Luminosity depends on the surface area of star $(4\pi R^2)$

Here, it is given that the two stars are at the same distance from the Earth. The first star is dimmer. We can conclude from this that first star star must be smaller than the second star.

7 0

3 years ago

Read 2 more answers