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Fittoniya [83]
3 years ago
5

A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel

ocity in meters/second immediately after the collision?
Physics
1 answer:
Vesna [10]3 years ago
3 0

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

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If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
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Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

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