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nignag [31]
3 years ago
6

What are some of the main types of force? Explain.

Physics
2 answers:
Shalnov [3]3 years ago
5 0
Answer:
1. applied force
2. gravitational force
3. normal force
4. frictional force
5. air resistance force
6. tension force
7. spring force
8. electrical force
9. magnetic force
LiRa [457]3 years ago
4 0
Applied force is this answer
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8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow
Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

8 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

8 0
3 years ago
Read 2 more answers
A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to
leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s

So, the combined velocity is 8.61 m/s.

5 0
2 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
On a winter day with a temperature of -10°C, 500g of snow (water ice) is brought inside where the temperature is 18 °C. The snow
Serjik [45]

Answer:

Explanation:

Mass of ice m = 500g = .5 kg

Heat required to raise the temperature of ice by 10 degree

= mass of ice x specific heat of ice x change in temperature

= .5 x 2093 x 10 J

10465 J

Heat required to melt the ice

= mass of ice x latent heat

0.5 x 334 x 10³ J

167000 J

Heat required to raise its temperature to 18 degree

= mass x specific heat of water x rise in temperature

= .5 x 4182 x 18

=37638 J

Total heat

=10465 +167000+ 37638

=215103 J

7 0
3 years ago
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