Answer:

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom? b) Calculate the flow speed in the bathroom. c) What is algebraic expression for the pressure in the bathroom? d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

A beam of light, which is traveling in air, is reflected by a glass surface. Does the reflected beam experience a phase change, and if so, by how much is the phase of the beam changed?

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

**Answer:**

y = 80.2 mille

**Explanation:**

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a **telescope**, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum **distance** at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered **telescope** would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

#SPJ3

The distance for both Parts A and B are given in the question.

A balloon drifts **140m** toward the west in 45s.

The wind suddenly changes and the balloon flies **90m** toward the east in the next 25s.

To find the total distance, we can just add.

140 + 90 = **230m**

Best of Luck!

**Answer:**

Gamma rays, x-rays, visible light, infrared radiation and radiowaves

**Explanation:**

Gamma rays, x-rays, ultraviolet, visible light, infrared radiation, microwave and radiowaves

b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?

Express your answer with the appropriate units.

**Answer:**

6052114.67492 m

**Explanation:**

v = Velocity of cosmic ray = 0.88c

c = Speed of light =

d = Width of Earth = Diameter of Earth =

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

**The Earth's width is 6052114.67492 m**

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

**Hence, the width is**

Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

To determine the mass of the second cart and its speed after impact, we can use the principle of conservation of momentum. The initial momentum of the first cart is equal to its final momentum plus the momentum of the second cart. After calculating the mass of the second cart, we can use the conservation of momentum again to find its speed by equating the final** velocity** of the combined carts to the initial velocity of the first cart.

To determine the mass of the second cart, we can use the principle of conservation of momentum. The initial momentum of the first cart, with a mass of 340 g and an initial velocity of 1.2 m/s, is equal to its final momentum plus the momentum of the second cart. Using this equation, we can solve for the mass of the second cart.

After calculating the mass of the second cart, we can use the conservation of **momentum** again to find its speed after the impact. Since the two carts stick together after the collision, the final velocity of the combined carts is equal to the initial velocity of the first cart. Using this equation, we can solve for the speed of the second cart.

#SPJ11

**Answer:**

the correct option is C

**Explanation:**

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C