What medium do seismic waves travel through?

Two states of matter are described below. State A: Cannot be compressed and retains its shape State B: Highly compressible Which

telo118 [61]

Answer:

State A = piece of metal; State B = air

Explanation:

For the three main states of matter here's how it breaks down.

Solid - Cannot be compressed and retains its shape

Liquid - Cannot be compressed and does not retain its shape

Gas - Compressible and does not retain its shape.

Knowing this State A has to be solid. Only one of the options has A as a solid, so that's the answer. Worth knowing state B is a gas though, only one compressible, just like solid is the only one that retains its shape.

7 0

3 years ago

Read 2 more answers

A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr

aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0

2 years ago

If the moon is a full moon tonight what will be the moon one week later waxing or waning

tatiyna

-- Starting from nothing (New Moon), the moon's shape grows ('waxes')
for half of the cycle, until it's full, and then it shrinks ('wanes') for the next
half of the cycle.

-- The moon's complete cycle of phases runs 29.53 days . . . roughly
four weeks.

-- So, beginning from New Moon, it spends about two weeks waxing until
it's full, and then another two weeks waning until it's all gone again.

-- After a Full Moon, the moon is waning for the next two weeks. So it's
definitely waning at one week after Full.

5 0

2 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

3 years ago

A runner moves west with an initial velocity of 4 m/s. She accelerates 3 m/s2 for 30 seconds. The runner's final velocity is

Firlakuza [10]

Answer:

v = 94m/s

Explanation:

Using the first equation of motion

v = u + at

u = 4m/s , a = 3m/s² , t = 30s , v = ?

v = u + at

v = 4 + 3 × 30

v = 4 + 90

v = 94m/s

I hope this was helpful, please mark as brainliest

3 0

3 years ago