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lesya [120]
2 years ago
6

A 3600-N Force causes a car to accelerate at a rate of 4m/s2. What is the mass of the car?

Physics
1 answer:
Mademuasel [1]2 years ago
3 0

Answer:

14400 kg

Explanation:

According to Newton's second law of motion, F = ma

F = 3600 N

a = 4m/s2

3600 = m/4

3600*4 = m

m = 14400 kg

Therefore the mass of the car is 14400 kg

Hope u understand

Please mark as brainliest

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Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp
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Answer:

a=\frac{mBg-mAgSin\theta}{mA+mB}

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Given two mass on an incline code mA and mB and an angle of inclination \theta. g. Assume that mA is the weight being pulled up and mB the hanging weight.

-The equations of motion from Newton's Second Law are:

mBg-T=mBa where a is the acceleration.

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mBg-mAsin\theta-mAa=mBa

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At what position or positions on the x-axis is the electric field zero?
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Answer:

The electric field will be zero at x = ± ∞.

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Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

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E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

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\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

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\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

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