A series circuit has a 12-volt power source and two resistors of 1 ohm and 2 ohms respectively. How many amps will flow when

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

6 0

3 years ago

How do molecules at warm temperatures differ from molecules at cool temperatures? Question 1 options: At warm temperatures, mole

yaroslaw [1]

Answer:

A. At warm tempetures, molecules move around more.

Explanation:

I'm at k12 and I just took the test got it right. Physical Science: Unit 2 Test

7 0

3 years ago

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A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?

Bogdan [553]

The boy's momentum is 160 kg*m/s north.

The formula of momentum is p = mv, where p is momentum.
p = 40 kg * 4m/s north
p =160 kg*m/s north

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4 0

3 years ago

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If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp

Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

$m_1(v_f - v_i) = m_2(\Delta v)$

given that

$m_1 = 5 ounce$

$m_2 = 32 ounce$

For ball the change in speed will be

$v_f - v_i = (120 - 77)mph$

now from above equation

$5\times (120 - 77) = 32 \times \Delta v$

$\Delta v = 6.72 mph$

so speed of bat will decrease by 6.72 mph

3 0

3 years ago

In the diagram, R1 = 40.0 ,

Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0

3 years ago