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2 years ago

WEATHER: What kind of weather occurs along this type of front?

Physics

2 answers:

Radda [10]2 years ago

4 0

Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis

konstantin123 [22]2 years ago

3 0

Answer:

Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis

Explanation:

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A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

Dvinal [7]

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

$s=0+\0.5\times 21\times (2)^2$

$s=42\ m$

so Jet Plane travels a distance of 42 m in 2 s

5 0

3 years ago

What is the angular displacement of a minute hand of a clock after 3 minutes?

Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0

1 year ago

A ball traveling with an initial momentum of 5.1 kgm/s bounces off a wall and comes back in the opposite direction with a moment

Elina [12.6K]

Answer: Change in momentum=9.4kgm/s

Impulse=9.4kgm/s

Explanation:

Change in momentum=5.1-(-4.3)=5.1+4.3=9.4kgm/s

Impulse=Change in momentum

There impulse=9.4kgm/s

4 0

3 years ago

What will occur when the trough of Wave A overlaps the trough of Wave B?

Pavel [41]

C because the wave length was longer

6 0

3 years ago

Read 2 more answers

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f

Butoxors [25]

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = $\frac{p}{2l} \sqrt{\frac{T}{m} }$

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 = $\frac{3}{120} \sqrt{\frac{T}{2} }$ I

Now if there is 5 antinodes , the value of p = 5

Thus n = $\frac{5}{120} \sqrt{\frac{T}{2} }$ II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

3 0

2 years ago