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vaieri [72.5K]
3 years ago
14

a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1

6N )and f2 =-(12.0N)+(8.0N) find the third force
Physics
1 answer:
tekilochka [14]3 years ago
8 0

By Newton's second law, the net force on the object is

∑ <em>F</em> = <em>m</em> <em>a</em>

∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N

Let <em>f</em> be the unknown force. Then

∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>

=>   <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N

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In a collision experiment, the ratio of the velocity change between two carts of equal inertia is found to equal 1.
VashaNatasha [74]

Answer:

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

Explanation:

A vehicle crash problem must be solved with the equation of the moment,

Initial instant Before crash

              p₀ = m v₁ + mv₂

After the crash

            p_{f} = m v_{1f} + m v_{2f}

           p₀ = p_{f}

If the speed ratio before and after the crash is one

           p₀ / p_{f} = 1

We can assume that initially one of the cars was stopped

           m v₁₀ = m v_{2f}

           v₁₀ = v_{2f}

For the two speeds to be equal, the masses of the vehicles must be the same.

A) Therefore if I double the masses with are in the two terrine they are simplified and the radii of the speeds remain the same

B) If the masses are maintained and the speeds are doubled, the radius of the two speeds remains the same

5 0
3 years ago
A 52-kg snow skier is at the top of a 245-m-high hill. After she has gone down a vertical distance of 112 m, what is her total e
Assoli18 [71]

Answer:

  a. 125 kJ

Explanation:

Her total energy is the same as the potential energy she had at the top of the hill:

  PE = mgh

  = (52 kg)(9.8 m/s^2)(245 m) = 124,852 J

  ≈ 125 kJ . . . . matches choice A

_____

After skiing down 112 m, some of her initial energy is converted to kinetic energy, and some remains as potential energy. We assume the ski slope is essentially frictionless, and air resistance is negligible.

7 0
3 years ago
A car accelerates uniformly from rest to 23 m/s over a distance of 30 meters. What is the acceleration of the car?
kolezko [41]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

3 0
3 years ago
What is the magnitude of fs on an object lying on a flat surface without moving, on
Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

  • magnitude of force on the object, F = 10 N
  • angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
loris [4]

Answer:

The acceleration of the crate is 1.82\ m/s^2.

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

F=ma\\\\F-f=ma

f is frictional force, f=\mu N=\mu mg

F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2

So, the acceleration of the crate is 1.82\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
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