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AlekseyPX
3 years ago
9

What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (cw = 4 186 J/kg⋅°C)

Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

                              = 10 x 60 = 600 s

using equation of specific heat  

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t =  m S ΔT

800 x 600  = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C

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Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
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v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

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v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

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3 years ago
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3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

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The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

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v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
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