A northfield raider defensive back runs 20 m in 2.5 s what’s the average speed

What are some types of landforms on Earth’s surface?<br><br><br><br> PLS ANSWER QUICK 11 POINTS

Brrunno [24]

Answer:

plateau, mountains, hills, plains

5 0

3 years ago

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If you were to travel from the equator to the higher latitudes (near either pole) on Saturn, the differential rotation would cau

Dmitrij [34]

Answer:

Saturn's differential rotation will cause the length of a day measures to be longer by 0.4 hours

Explanation:

Differential rotation occurs due to the difference in angular velocities of an object as we move along the latitude of the or as we move into different depth of the object, indicating the observed object is in a fluid form

Saturn made almost completely of gas and has differential motion given as follows

Rotation at the equator = 10 hours 14 minutes

Rotation at high altitude = 10 hours 38 minutes

Therefore;

The differential rotation = 10 hours 38 minutes - 10 hours 14 minutes

The differential rotation = 24 minutes = 24 minutes × 1 hour/(60 minutes) = 0.4 hours

The differential rotation = 0.4 hours

Therefore, the measured day at the higher altitude will be 0.4 longer than at the equator.

8 0

3 years ago

Select the correct answer.

spin [16.1K]

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

7 0

3 years ago

A new ride being built at an amusement park includes a vertical drop of 71.6 meters. Starting from rest, the ride vertically dro

Allisa [31]

Answer: 2.6x107

Explanation:

3 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago