In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the

Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ =  acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t²                  (1)
x = 0.5a₂t²                  (2)

F = m₁a₁ = m₂a₂         (3)
Additionally,
x + y = 14000 m          (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

Answer:

Please refer to the explanation

Explanation:

Race distance is 5km

Top speed = 45 yards

Converting yards to kilometers:

1km equals 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x indicates the maximum distance he can sustain his highest speed in kilometers.

Thus, from the data available, we can determine that Lamar will not be able to maintain his maximum speed for the full 5km race, as he can only sustain it for 0.0411 kilometers.

Answer:

The current flowing through the tube is 0.601 A

Explanation:

Given data;

the diameter of the fluorescent tube is d = 3 cm

the incoming negative charge in the tube is -e = 3 x 10¹⁸ electrons/second

the outgoing positive charge equals +e = 0.75 x 10¹⁸ electrons/ second

The current within the fluorescent tube results from both positive and negative charges which contribute to maintaining electrical neutrality in the conductor (fluorescent tube).

Q = It

I = Q/t

where;

I signifies current in Amperes (A)

Q represents charge measured in Coulombs (C)

t denotes time which is in seconds (s)

1 electron (e) accounts for 1.602 x 10⁻¹⁹ C

Thus, 3 x 10¹⁸ e/s can be computed as

= (3 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

= 0.4806 C/s

This reflects the negative charge per second (Q/t) = 0.4806 C/s

Meanwhile, the positive charge per second amounts to

(0.75 x 10¹⁸ e/s  x 1.602 x 10⁻¹⁹ C) / 1e

Thus, the positive charge per second is equal to 0.12015 C/s

The total charge per second in the tube is then obtained by summing both charges: Q / t = (0.4806 C/s + 0.12015 C/s)

                                                                I = 0.601 A

Therefore, the current within the tube computes to 0.601 A

According to the second law, heat, often called thermal energy, cannot be entirely turned into work.
The second statement is closely tied to this law.
We can conclude that some energy dissipates while some is used for work.

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