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2 years ago

Drag the tiles to the correct boxes to complete the pairs.

Physics

1 answer:

marishachu [46]2 years ago

7 0

Answer:

Air pollution---> smog

Water pollution---> eutrophication

Land pollution---> contaminated soil

Light pollution---> sky glow

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As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0

Alika [10]

velocity of the physics instructor with respect to bus

$v = 6 m/s$

acceleration of the bus is given as

$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

$a = -2 m/s^2$

now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance

7 0

2 years ago

A student walks 3 blocks east, 4 blocks north, and 3 blocks west. What is the displacement of the student?

12345 [234]

4 blocks north because he is it not asking for north east

5 0

2 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = $\frac{KQ^{2} }{I^{2} }$

The direction of the force on charge 2Q is calculated as

tan ∅ = $\frac{f_{y} }{f_{x} }$ = 1.8284

therefore ∅ = $tan^{-1}$ 1.8284

= 61⁰

3 0

2 years ago

Will give Brainliest!! Question attached.

ss7ja [257]

Answer:

im pretty sure it is 3.0 K

Explanation:

8 0

2 years ago

Read 2 more answers

A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,

iris [78.8K]

Answer:

The work done by gravity is $4.975 \: Joules$

Explanation:

The data given in the question is :

Mass is $0.245 kg$

Height from ground is $2.07 m$

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = $mass \times gravity \times (change \: in \: height)$

Work done = $0.245 kg \times 9.81 m/s^{2} \times 2.07 m$

So, work done = $4.975 Joules$

Hence the work done by gravity is $4.975 \: Joules$

3 0

3 years ago