Hat indoor air pollutant often seeps up through cracks in the floor?

Using a cathode ray tube, Thomson confirmed that

Natasha2012 [34]

atoms are made of particles that have a negative charge.

4 0

4 years ago

An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10

Akimi4 [234]

Answer:

a) $F=475.7Hz$

b) $F'=410.899Hz$

Explanation:

From the question we are told that:

Velocity of eagle $V_1=35m/s$

Frequency of eagle $F_1=440Hz$

Velocity of Black bird $V_2=10m/s$

Speed of sound $s=343m/s$

a)

Generally the equation for Frequency is mathematically given by

$F=f_0(\frac{v-v_2}{v-v_1})$

$F=440(\frac{343-10}{343-35})$

$F=475.7Hz$

b)

Generally the equation for Frequency is mathematically given by

$F'=f_0(\frac{v+v_2}{v+v_1})$

$F'=440(\frac{343+10}{343+35})$

$F'=410.899Hz$

7 0

3 years ago

An organism that cannot make its own food called

Rzqust [24]

Answer:

heterotroph

Explanation:

8 0

3 years ago

Read 2 more answers

When plugging in metric facts, always remember that 1 big unit = # small units. Fill in these facts: 1.________ s= ________μs 2.

balandron [24]

Answer:

1. 1 s = 1 x 10⁶ μs

2. 1 g = 0.001 kg

3. 1 km = 1000 m

4. 1 mm = 1 x 10⁻³ m

5. 1 mL = 1 x 10⁻³ L

6. 1 g = 100 dg

7. 1 cm = 1 x 10⁻² m

8. 1 ms = 1 x 10⁻³ s

Explanation:

1.

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

1 s = 1 x 10⁶ μs

2.

1000 g = 1 kg

1 g = 1/1000 kg

1 g = 0.001 kg

3.

1 km = 1000 m

4.

1 mm = 1 x 10⁻³ m

5.

1 mL = 1 x 10⁻³ L

6.

1 x 10⁻² g = 1 dg

(1 x 10⁻² x 10²) g = 1 x 10² dg

1 g = 100 dg

7.

1 cm = 1 x 10⁻² m

8.

1 ms = 1 x 10⁻³ s

4 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers