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solmaris [256]
3 years ago
6

CAN YOU SUBSCRIBE? PLEASE...

Physics
1 answer:
nadezda [96]3 years ago
4 0

Answer:

just wait I'll subscribe yours

<h2>hope it helps you have a good day keep smiling be happy stay safe</h2>

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How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
Why is pluto not a planet?
mihalych1998 [28]
Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.

Hope this helps!
3 0
3 years ago
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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
3 years ago
A box weighing 10 N is 4 m away from the fulcrum. How much force must be applied 1 m away on the opposite end of the fulcrum in
julia-pushkina [17]

Answer: Socratic

Explanation: i’m not sure but you can use Socratic it’s a good app that helps

4 0
3 years ago
An airplane pilot sets a compass course due west and maintains a speed of 220 km/h. after flying for a half hour she finds herse
timurjin [86]

The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.  

so airspeed due west is  = net speed - speed of plane = 240-220= 20 km/h.  

airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.  

the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h

the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.  

if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.

so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.  

the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.

5 0
2 years ago
Read 2 more answers
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