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aleksandr82 [10.1K]
3 years ago
7

Why might be people be causing global warming?

Physics
1 answer:
amid [387]3 years ago
8 0
Pollution? 
:):):):):):)
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A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

6 0
3 years ago
What is the function of the phytoplankton in ocean ecosystems?
vesna_86 [32]
Ok ok! It's A. Producer
5 0
3 years ago
friction between the tire of a moving car and the dry pavement is: A. static B. rolling C. sliding D. riding
qwelly [4]

Answer:A static

Explanation:

6 0
2 years ago
What is the difference between landforms and landmarks
Anika [276]
A land form is a natural feature of earths surface and a land mark is can be man-made or natural and it is used to mark something significant. 
7 0
3 years ago
A force of 6 N will stretch a rubber band 4 cm ​(0.04 ​m). Assuming that​ Hooke's law​ applies, how far will an 18​-N force stre
Bogdan [553]

Explanation:

First, we need to calculate the constant force, that is, the ratio between the applied force and the rubber stretch due to the application of the force:

k=\frac{F}{d}(1)\\k=\frac{6N}{0.04m}\\k=150\frac{N}{m}

Now, we can know how far will an 18​N force stretch the rubber​. From (1):

d=\frac{F}{k}\\d=\frac{18N}{150\frac{N}{m}}\\d=0.12m=12cm

The work done by the external force on the rubber is equal to its elastic potential energy:

W=\frac{kd^2}{2}\\W=\frac{150\frac{N}{m}(0.12m)^2}{2}\\W=1.08J

3 0
3 years ago
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