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Rashid [163]
2 years ago
15

Which of the following describe wind patterns created by a low-pressure front, where two air masses are interacting? Choose one

answer below.

Physics
1 answer:
viva [34]2 years ago
6 0

Answer:

I think its  might Be number B

Explanation:

tell me if im rong

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Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
Burka [1]
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
7 0
2 years ago
Read 2 more answers
A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh
DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force =\mu \left ( N\right )=0.4\times \left ( 740+160\right )

=360 N

cos\theta =\frac{3}{5}

sin\theta =\frac{4}{5}

(b)Moment about Ground Point

740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

N_1tan\theta =1140

N_1=171 N

N_1=f=171 N

(c)

740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta

Here maximum friction force can be 360 N

Therefore N_1=360 N

Where x is the maximum distance moved by man along the ladder

360\times 5\times \frac{4}{3}=740x+160\times 2.5

740x=2000

x=2.702m

5 0
2 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

5 0
3 years ago
Time dilation means that
Anastaziya [24]
The answer is D. time really does pass more slowly in a rest frame of reference relative to a frame of reference that is moving
5 0
3 years ago
An interplanetary spacecraft is moving at
mixer [17]
The awnser is. 1728000 kilometers
7 0
3 years ago
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