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1 year ago

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Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A

cart B of mass m2 = 4 kg, initially at rest. The two carts combine together and move as one object. Calculate the height reached
by the two carts after collision.

1 answer:

Harrizon [31]1 year ago

3 0

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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4 0

3 years ago

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Gekata [30.6K]

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = $6.5*10^{4}\ m/s$

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity $[\theta]$ = 15 degree.

We are asked to calculate the magnetic force experienced by the charged particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force $\vec F=q(\ \vec V \times \vec B)$

i.e F = $=qVBsin\theta$

$=6.8*6.5*10^{4}*1.4*sin15$

$=61.88*10^{4}sin15$

$=61.88*10^4*0.2588\ N$

$=16.016*10^4\ N$

$=1.6*10^5\ N$

Hence, the force experienced by the charged particle is C i.e $1.6*10^5 N$

4 0

2 years ago

Read 2 more answers

What voltage would be measured across the 45 ohm resistor?

Tamiku [17]

You would find the potential difference aka voltage, but more specifically it would be just the voltage that the resistor uses and not the whole circuit.

But if you want the voltage value it’s V=IR so whatever the current is multiply it by the 45 ohm resistor value

8 0

3 years ago

Read 2 more answers

A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c

sergiy2304 [10]

Answer:

Part a)

$a_{hanger} = g - \frac{T}{m}$

Part b)

$a_{cart} = \frac{T}{M}$

Part c)

$a = \frac{mg}{M + m}$

Part d)

$a = 1.35 m/s^2$

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

$mg - T = ma$

so we have

$a_{hanger} = g - \frac{T}{m}$

Part b)

now for car that is rolling on the floor the net force is given as

$F = Ma$

$T = Ma$

$a_{cart} = \frac{T}{M}$

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

$T = Ma$

$mg - Ma = ma$

$a = \frac{mg}{M + m}$

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

$a = \frac{0.50(9.81)}{2.40 + 0.50}$

$a = 1.35 m/s^2$

6 0

3 years ago

2. A person lifts 200kg seven times over the course of 11.8s. If they displaced the weight 2.2m up each time, how much power did

Aneli [31]

Answer:

<em>The person delivered a power of 2,558 Watt</em>

Explanation:

<u>Work and Power</u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being the force vector and the displacement vector, the work is calculated as:

$W=\vec F\cdot \vec s$

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred per unit of time. In the SI, the unit of power is the watt, equivalent to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

If the person lifts a mass of m=200 Kg, then exerts a force equal to its weight:

F = m.g = 200*9.8 = 1,960

F = 1,960 N

The work done when lifting the weight 7 times by a distance of s=2.2 m is:

W = 7*1,960*2.2=30,184

W = 30,184 J

Finally, the power delivered in t=11.8 seconds is:

$\displaystyle P=\frac {30,184}{11.8}$

P = 2,558 Watt

The person delivered a power of 2,558 Watt

4 0

3 years ago