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3 years ago

Help needed

Physics

1 answer:

MakcuM [25]3 years ago

6 0

Answer: Atmosphere and geosphere.

Geo-sphere is the solid part of the earth. Hydrosphere is the water part. The living things on the earth make up the biosphere. The gases, water vapors etc make up the atmosphere. When a volcanic eruption adds carbon dioxide to the air, the carbon-dioxide is being added from geosphere to the atmosphere. Hence, there is interaction between atmosphere and geosphere.

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Find the voltage across the 15 Ohm resistor plz help ty

jonny [76]

Answer:

5.4 Volts is the answer!

Explanation:

Good luck with Acellus!

7 0

2 years ago

A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear

vichka [17]

Answer:

a) λ = 5.19 10⁻⁴ C/m , b) E = 1,573 10⁻³ N/C , c) the direction of the field is directed to the bar

Explanation:

a) the linear density defined as the ratio between the charge per unit length

λ = q / l

Let's start by reducing the units to the SI system

L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

a = 12 cm (1m / 100cm) = 12 10⁻² m

q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

λ = 5.19 10⁻⁴ C/m

b) Let's look for the electric field for a point at a distance a from the end of the bar

E = k dq / r²

To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

λ = dq/dx

dq = λ dx

E = k ∫ λ dx / x²

We integrate and evaluate between the lower limits x = a and higher x = L + a. Here we place the test point at the origin of the system

E = k λ (-1 / x)

E = k λ (-1 /(L + a) + 1 /a)

E = k λ (L /a(L + a)

Let's change the density for its value

E = k (q / L) (L / a (L + a)

E = k q 1 /[a(L + a)]

E = 8.99 10⁹ 4.23 10⁻¹⁵ [1 /12 10⁻²(8.15 10⁻² + 12 10⁻²)]

E = 1,573 10⁻³ N/C

c) the direction of the field is directed to the bar, because it has a negative charge

d) now we change the distance a = 50 cm = 0.50 m

Bar

E = 8.99 10⁹ 4.23 10⁻¹⁵ ( 1 /0.5(0.0815 +0.5))

E = 1,308 10⁻⁴ N/C

Charge point

q = -4.23 10⁻¹⁵ C

E = k q / r²

E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

E = 1.521 10⁻⁴ N/C

7 0

3 years ago

Lightning results from ________.

just olya [345]

An imbalance between electrical charges

8 0

2 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

$H =212.6 \ J$

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$