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2 years ago

A 9.5 kg object undergoes an acceleration of 3.3m/s^2. What is the magnitude of the next external force acting on it

Physics

1 answer:

Black_prince [1.1K]2 years ago

5 0

The net external force is 31.4 N

Explanation:

We can solve this problem by applying Newton's second law, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the object in this problem, we have:

m = 9.5 kg is the mass

$a=3.3 m/s^2$ is the acceleration

Solving for F, we find the net force:

$F=(9.5)(3.3)=31.4 N$

Learn more about Newton's second law:

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Thanks + BRAINLIST <br><br> Please need correct answer asappp

nirvana33 [79]

Answer:

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2 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

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Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

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the y comonent of the vecor is Ay = 49.92 N

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2 years ago

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

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2 years ago