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dybincka [34]
3 years ago
14

On a cross-country trip, a couple drives 500 mi in 10 h on the first day, 380 mi in 8.0 h on the second day, and 600 mi in 15 h

on the third day. what was the average speed for the whole trip?
Physics
1 answer:
Zepler [3.9K]3 years ago
8 0

We know that the average speed is simply the ratio of the total distance travelled over the total duration of the trip.

total distance = 500 mi + 380 mi + 600 mi

total distance = 1,480 mi

 

total time = 10 h + 8 h + 15 h

total time = 33 h

 

So the average speed is therefore:

average speed = 1,480 mi / 33 h

<span>average speed = 44.85 mi / h</span>

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A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

3 0
3 years ago
The block in the drawing has dimensions L0×2L0×3L0, where L0 =0.5 m. The block has a thermal conductivity of 250 J/(s·m·C˚). In
stira [4]

Answer:

I am sorry but what drawing

3 0
2 years ago
Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As
Troyanec [42]

Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

Given

Mass = 45000kg

acceleration = 70m/s^2

Substitute

F = 45000 * 70

F = 3,150,000N

Hence the force required to be produced by the rocket engines is 3,150,000N

8 0
3 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
2 years ago
NEED HELP TODAY
Free_Kalibri [48]

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

6 0
2 years ago
Read 2 more answers
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