3 years ago

Help me with this please

Physics

1 answer:

Tanya [424]3 years ago

8 0

Answer:

century eh i

Explanation:

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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

Viefleur [7K]

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

6 0

3 years ago

A light spring has a force constant of 70 N/m and is used to pull a 10 kg box on a horizontal frictionless surface. If the box h

Sergio039 [100]

Answer:

82.4 cm

Explanation:

∑F = ma

kx cos θ = ma

x = ma / (k cos θ)

x = (10 kg) (5 m/s²) / (70 N/m cos 30.0°)

x = 0.824 m

x = 82.4 cm

8 0

3 years ago

Find the acceleration of a car with the mass of 1,200 kg and a force of

Mashutka [201]

Answer:9.17 m/s^2

Explanation:

mass=1200kg

Force=11 x 10^3 N

Acceleration=force ➗ mass

Acceleration=11 x 10^3 ➗ 1200

Acceleration=9.17

Acceleration=9.17 m/s^2

3 0

2 years ago

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

3 years ago

When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav

GuDViN [60]

I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.

4 0

3 years ago