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Illusion [34]
3 years ago
9

Is any of this right?

Physics
1 answer:
Amiraneli [1.4K]3 years ago
5 0
You will need to multiply them.
I have physics class and for acceleration we use 9.8m/s^2 but you used 9.5m/s^2.
If that’s what ur teacher wants u to use that’s ok. But yeah you multiply them.
F=ma
You multiply m and a
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luda_lava [24]
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
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8 0
3 years ago
How does the loneliest whale relate to physics/waves.
Lelechka [254]
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What is sex And new born baby why connected to women's veins?
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3 0
3 years ago
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
Deffense [45]
The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry
7 0
3 years ago
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