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2 years ago

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Physics Physics Physics - 10

1 answer:

Svetach [21]2 years ago

7 0

Answer:

1.Pulley

2. Wheel and axle

3. Screw

4. wedge

5. Incline

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

Explanation's of E=MC²

natita [175]

Answer:

E means energy

M= Mass

C=speed of light squared (the exponent means squared)

4 0

2 years ago

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A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature $\left ( T_H\right )=240^{\circ}\approx 513$

Lower Temprature $\left ( T_L\right )=20^{\circ}\approx 293$

Engine power ouput $\left ( W\right )=910 W$

Efficiency of carnot cycle is given by

$\eta =1-\frac{T_L}{T_H}$

$\eta =\frac{W_s}{Q_s}$

$1-\frac{293}{513}=\frac{910}{Q_s}$

$Q_s=2121.954 W$

$Q_r=1211.954 W$

rounding off to two significant figures

$Q_r=1200 W$

5 0

2 years ago

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

Em₀ = U = mg y

Final. Low point just before the crash

Emf = K = ½ m v²

Em₀ = Emf

m g y = ½ m v²

v = √ 2 g y

Let's calculate

v = √ (2 9.8 0.05)

v = 0.99 m / s

1) the moment before the crash is

p₀ = m v

p₀ = 0.221 0.99

p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

p = 0

2) change of momentum

Δp = p - p₀

Δp = 0- 0.219

Δp = -0.219 kg m / s

3) the reason is

Δp / p = 1

In percentage form it is 100%

3 0

3 years ago

Freeeeeeeeeeeeeeeeeeeeeeeee points

andreyandreev [35.5K]

Answer:

yehyehhfujghnnkoohu

3 0

2 years ago

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