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2 years ago

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Two adjacent students stand on stationary skateboards, face each other, and push apart. The skateboarder on the left weighs 65 k

g, the one on the right is 85 kg. If the lighter skateboarder moves at 3.2 m/s, determine the speed and direction of the heavier skateboarder.

1 answer:

grigory [225]2 years ago

4 0

As we know that two skateboarder will push each other so there is no external force on this system

Hence we can use momentum conservation here

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0 + 0 = (65)(3.2) + 85(v)$

now we have

$0 = 208 + 85 v$

$v = - 2.45 m/s$

so the heavy skateboarder will move off with speed 2.45 m/s in the opposite direction

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Janelle stands on a balcony, two stories above Michael. She throws one ball straight up and one ball straight down, but both wit

antoniya [11.8K]

Answer:

Both balls have the same speed.

Explanation:

Janelle throws the two balls from the same height, with the same speed. Both balls will have the same potential and kinetic energy. Energy must be conserved. When the balls pass Michael, again they must have the same potential and kinetic energy.

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What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?

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Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a = the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
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Answer: 0.2 m/s²

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2 years ago

A battery with internal resistance r is connected to a load resistance R. If R is increased, does the terminal voltage of the ba

Vitek1552 [10]

Answer:

The terminal voltage of the battery decreases.

Explanation:

An idea battery does not have internal resistance but a real practical battery always have an internal resistance r connected in series with the battery.

This internal resistance causes a voltage drop when load R is connected, a current I flows in the circuit which causes a voltage drop I*r in the battery therefore, the terminal voltage of the battery will decrease.

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2 years ago

A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

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2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

butalik [34]

Answer:

The temperature of the gas is 1197.02 K

Explanation:

From ideal gas law;

PV = nRT

Where;

P is the pressure of the gas

V is the volume of the gas

R is ideal gas constant = 8.314 L.kPa/mol.K

T is the temperature of the gas

n is the number of moles of gas

Volume of the gas in the cylindrical container = πr²h

Given;

r = 6/2 = 3 cm = 0.03 m

h = 11 cm = 0.11 m

V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L

number of moles of oxygen gas = Reacting mass / molar mass

$=\frac{0.1}{32} = 0.003125, moles$

$T = \frac{PV}{nR} = \frac{100X0.311}{0.003125X8.314} =1197.02K$

Therefore, the temperature of the gas is 1197.02 K

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2 years ago