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2 years ago

A student is conducting an experiment to determine how far a ball will roll down a ramp based on the angle of incline.

Physics

2 answers:

Nataly_w [17]2 years ago

7 0

Answer:

Explanation:

There are several possible controls that can be made in order to control the experiment.

The angle of inclination

The size and mass of the ball.

The length of the ramp.

All these factors are controllable and very well define the experiment. More the weight and angle of inclination, faster and longer distance the ball will cover.

Sedbober [7]2 years ago

5 0

The size of the ball, the mass, and if the ball gets an extra push

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A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2

leva [86]

Answer:

Tension = 0.012 N

Explanation:

If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,

Tension = Weight

Tension = mg

where,

m = mass of spider = 1.27 g = 0.00127 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

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3 years ago

Which balances the equation Mg + O2 — MgO?

Stolb23 [73]

Answer:

To balance an equation such as Mg + O2 → MgO, the number of the atoms in the product must equal the number of the atoms in the reactant. Mg + O2 --> MgO. To balance an equation, we CAN change coefficients, but NOT SUBSCRIPTS to balance equations.

Explanation:

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2 years ago

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$