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3 years ago

In the lab, you performed a trial where the fan was turned

Physics

2 answers:

beks73 [17]3 years ago

7 0

Answer:

Simple Response The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

Ksju [112]3 years ago

6 0

Answer:

Sample Response: The first part of the graph is a straight diagonal line and the second part of the graph is a straight horizontal line. The slope of the first part of the graph is positive, indicating that the cart increased in speed, and was therefore accelerating for the first four seconds. After the fourth second, the slope of the line is zero, indicating that the cart moved at a constant speed and did not accelerate any more because the fan was turned off.

Explanation:

on edge :)

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A young man heaves a 6.5 kg rock at a velocity of 6.9 m/s. What is the kinetic energy of the rock?

earnstyle [38]

Answer:

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2 years ago

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

3 years ago

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north

Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

s ₁ = 1.93 i km

1.03 km due south

s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

Magnitude of displacement, $=\sqrt{(-0.39)^2+2.03^2}=2.07km$

Time taken = 1.771 hour

Magnitude of average velocity, $=\frac{2.07}{1.771}=1.17km/hr$

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3 years ago

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lyudmila [28]

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OlgaM077 [116]

Answer: Moral

Explanation:

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3 years ago