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Anna35 [415]
3 years ago
9

An air track car with a mass of 0.55 kg and velocity of 5.8 m/s to the right collides and couples with a 0.45 kg car moving to t

he left with a velocity of 3.9 m/s. What is the combined velocity of the cars just after the collision?
Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Explanation:

Given:

m1 = 0.55 kg

m2 = 0.45 kg

vi1 = +5.8 m/a

vi2 = -3.9 m/a

Solve for V

m1vi1 + m2vi2 = (m1 + m2)V

(0.55 kg)(+5.8 m/s) + (0.45 kg)(-3.9 m/s) = (0.55 + 0.45)V

V = +1.46 m/a

This means that the combined mass after the collision is moving to the right at 1.46 m/s

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Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J
kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

V=\dfrac{W}{q}

V=\dfrac{12}{0.0001}

V=12\times 10^4\ Volt

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

V=\dfrac{W}{q}

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3 years ago
the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds
ki77a [65]

Answer:

Explanation:

The question is incomplete.

The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.

s= sin2(pi)t

Acceleration = d²S/dt²

dS/dt = 2πcos2πt

d²S/dt² = -4π²sin2πt

A(t) = -4π²sin2πt

Next is to find acceleration after 4.5 seconds

A(4.5) = -4π²sin2π(4.5)

A(4.5) = -4π²sin9π

A(4.5) = -4π²sin1620

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3 years ago
what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?
zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

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<span>rocky surface

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