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mafiozo [28]
2 years ago
13

Calculate the voltage difference in a circuit with a resistance of 25 Ω if the current in the circuit is 0.5 A.

Physics
1 answer:
weeeeeb [17]2 years ago
8 0

Answer:

check the attachments for the solution

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The main difference between speed and velocity involves
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Speed is a scalar quantity, while Velocity is a vector quantity.
Speed is measured without taking into fact direction, but Velocity takes into fact the direction.
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An object that is not accelerating or decelerating has zero net force acting on it. Choose 1 answer: Choose 1 answer: A True B F
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Hey there,

Answer:

A, True

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3 years ago
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John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the
Mumz [18]

Answer:

30 minutes

Explanation:

Energy per time is constant, so:

E₁ / t₁ = E₂ / t₂

m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂

(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t

(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t

3/min = 90 / t

t = 30 min

6 0
3 years ago
Theo mixes cake batter and places it in the oven at 350°F for 45 minutes. Explain what type of reaction baking a cake represents
Nataly_w [17]

Baking a cake creates a endothermic reaction. When heated, the baking soda/powder reacts, the eggs change form, and everything else follows suit. The baking soda/powder creates the rising of the cake.

4 0
3 years ago
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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
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