Please help with both questions I’m giving all my points :) it’s 23 and 24

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

$= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}$

The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

8 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 108 m. Find

Goshia [24]

Answer:

It took 1.28 seconds to his voice to reach the Earth via radio waves.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

Light propagates as electromagnetic wave in vacuum with a speed of $3x10^{8}m/s$ . Therefore, radio waves will have in vacuum the same speed.

Then, to know the time that it took for its voice, the next equation can be used:

$c = \frac{d}{t}$ (1)

Where c is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to his voice to reach the Earth via radio waves.

7 0

3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0

3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago

Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistan

Rama09 [41]

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a) Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

3 0

3 years ago