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wariber [46]
2 years ago
8

4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat

Physics
1 answer:
nydimaria [60]2 years ago
3 0

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

4 0
2 years ago
How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
Rasek [7]

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

6 0
2 years ago
A boat takes off from the dock at 2.5 m/s and speeds up at 4.2 m/s2 for 6.0 s. How far has the boat traveled? Round your answer
Zinaida [17]
It's simple bro
91m is the answer

6 0
2 years ago
Read 2 more answers
The formula s = 16t2 gives the distance an object falls due to gravity, where s is the distance in feet and t is the time in sec
Georgia [21]
To determine the distance in units of feet of the object falling at a given time, we simply use the relation given above and substitute the given value of time to the function. We do as follows:

s = 16t^2
s = 16 ( 6 )^2
s = 576 feet

The distance of the object from the ground at 6 seconds is 576 ft.
8 0
2 years ago
A sharpening wheel is traveling at 5 rad/s, it slows down to rest in 30 seconds while sharpening an axe. What is its angular acc
Ratling [72]

Answer:

Angular acceleration = 0.167 rad/s^2

Explanation:

Given

Initial Angular velocity (w1) = 5 rad/s

Final Angular velocity (w2) = 0 rad/s

Time taken to change velocity from w1 to w2 = 30 seconds

Angular acceleration is equal to the change in angular velocity to the time taken for making thing change

Hence, Angular acceleration

\frac{w_2 -w_1}{t} \\\frac{5-0}{30}\\0.167rad/s^2

3 0
2 years ago
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