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2 years ago

When I mix 200g of water at 10°C with 300g of water at 55°C, what is the final temperature of the mixture?

Physics

1 answer:

Anarel [89]2 years ago

7 0

Answer:

4th of August is the

Explanation:

5th and then I can help out by getting

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The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s

Ronch [10]

KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J

4 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?

blsea [12.9K]

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T = 2541.799 K

The temperature = 2541.799 K.

7 0

3 years ago

Explain how heat (thermal energy) can change the motion (kinetic energy) of the particles in objects.

Vika [28.1K]

Answer:

Adding heat makes the particles move faster so the particles have more kinetic energy when more thermal energy is added

Explanation:

4 0

3 years ago

When I mix 200g of water at 10°C with 300g of water at 55°C, what is the final temperature of the mixture?​

When I mix 200g of water at 10°C with 300g of water at 55°C, what is the final temperature of the mixture?