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sleet_krkn [62]
2 years ago
10

Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of

Physics
1 answer:
rosijanka [135]2 years ago
7 0

Answer:

8.35 × 10^{-3} kg

Explanation:

Remember the formula for heat:

Q=mc\Delta T

You just need to solve for m, doing so you're left with:

m=\dfrac{Q}{c\Delta T}

Replace the variables with numbers and that's it.

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3 years ago
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

     v=\sqrt{\frac{T}{(m/l)}}

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

(m/l)=\frac{10}{V^2}

Explanation:

From the question we are told that

Speed of a transverse wave given by

v=\sqrt{\frac{T}{(m/l)}}

Maximum Tension is T=10.0N

Generally making (m/l) subject from the equation mathematically we have

v=\sqrt{\frac{T}{(m/l)}}

v^2=\frac{T}{(m/l)}

(m/l)=\frac{T}{V^2}

(m/l)=\frac{10}{V^2}

Therefore the Linear mass in terms of Velocity is given by

(m/l)=\frac{10}{V^2}

8 0
3 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
2 years ago
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