All categories

2 years ago

Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?

Physics

1 answer:

den301095 [7]2 years ago

6 0

Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

You might be interested in

A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.

aivan3 [116]

Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

F =20.0N

Theta = 37°

W = FSCos(theta)

W = 20×6.0×Cos(37)

W = 95.8J

8 0

2 years ago

Read 2 more answers

Suppose a current flows through a copper wire. Which two things occur?

worty [1.4K]

I’m pretty sure the answer is c and d hope this helps and good luck

3 0

2 years ago

Read 2 more answers

How do you write a hypothesis

Mashcka [7]

It's and if, then statement!

5 0

2 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

2 years ago

A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s

belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, $\omega_i=114\ rev/min = 11.93\ rad/s$

Final angular speed of the wheel, $\omega_f=0$

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

$I=mr^2$

$I=49\times (0.73)^2$

$I=26.11\ kg-m^2$

We know that the work done is equal to change in kinetic energy.

$W=\Delta E$

$W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

$W=-\dfrac{1}{2}\times 26.11\times (11.93^2)$

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{1858.05\ J}{22\ s}$

P =84.45 watts

Hence, this is the required solution.

4 0

2 years ago