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3 years ago

A narrow region between two air masses of different densities is a

Physics

1 answer:

Sav [38]3 years ago

5 0

A front is a narrow region between two air masses of different densities.

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What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the

alisha [4.7K]

Answer:

1.1648×10⁻¹¹ N

Explanation:

Using

F = qvBsinФ..................... Equation 1

Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.

Note: The angle between v and B = 90°

Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°

Substitute into equation 1

F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°

F = 11.648×10⁻¹²

F = 1.1648×10⁻¹¹ N.

6 0

3 years ago

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HELP PLZ "25 meters per second Northeast" is a description of

Hoochie [10]

Answer:

ion no

Explanation:

cuz i dont

7 0

3 years ago

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The electric current leaves the battery through the --- complete this blank

aev [14]

The bulb

Hope this helps!

5 0

4 years ago

A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s

Andrej [43]

Answer:

$1300 m$

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from $20 s$ to $70 s$ is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area $=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)$

$=200+1000+100$

$=1300$

As velocity is in $m/s$ and time is in $s$ so the unit of area is $m$

Hence, the total distance is $1300m$ .

8 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.83 times a second. A tack is stuck in the tire a

timama [110]

Answer:

The tangential speed of the tack is 6.988 meters per second.

Explanation:

The tangential speed experimented by the tack ( $v$ ), measured in meters per second, is equal to the product of the angular speed of the wheel ( $\omega$ ), measured in radians per second, and the distance of the tack respect to the rotation axis ( $R$ ), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second: