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4 years ago

What are the answers please help

Physics

1 answer:

aleksandr82 [10.1K]4 years ago

3 0

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.) 10.31 cm

e.) 10.24 cm

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.) If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm

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4 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignor

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Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

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where:

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time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

$v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}$

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

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This leaves only one unknown: v₀

$v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}$

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

$t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2$

and how fast will it go:

$v' = -gt + v_0$

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

$0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2$

Now let's use this result in the equation for the first ball:

$0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m$

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