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melisa1 [442]
3 years ago
11

If you change the mass of a moving object, you change its ??

Physics
1 answer:
satela [25.4K]3 years ago
7 0
If you change the mass, you change anything that depends on mass. But I suspect the word you're looking for is "momentum"
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Answer:

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Explanation:

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3 years ago
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A house is maintained at 1 atm and 24°C. Outdoor air at 5.5°C infiltrates into the house through the cracks and warm air inside
Lunna [17]

Answer:

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Explanation:

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3 0
3 years ago
Un móvil se desplaza con movimiento uniforme con una rapidez de 36 Km/h ¿Cuál es la distancia recorrida al cabo de 0,5 horas?
Diano4ka-milaya [45]

Answer:

Distancia = 17,5 kilómetros.

Explanation:

Dados los siguientes datos;

Velocidad = 36 km/h

Tiempo = 0.5 horas

Para encontrar la distancia recorrida;

Distancia = velocidad * tiempo

Distancia = 35 * 0.5

Distancia = 17,5 kilómetros.

Por tanto, la distancia recorrida por el automóvil es de 17,5 kilómetros.

4 0
3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
3 years ago
Questions 8 out of 20
MrMuchimi

Answer:

potential energy

Explanation:

8 0
3 years ago
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