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Kay [80]
3 years ago
5

with what force will the a car hit a tree if the car has a mass of 3,000 kg and a acceleration of 2m/s^2​

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
8 0

Answer:

<h3>The answer is 6000 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3000 × 2

We have the final answer as

<h3>6000 N</h3>

Hope this helps you

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3 years ago
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
Lelechka [254]

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ T_1Cos30^{\circ}-T_2Cos60^{\circ}=0

or,

    T_1Cos 30^{\circ}=T_2Cos60^{\circ}

On substituting the values, we get

      T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}

      T_1\times \sqrt{3} =T_2....(equation 1)

On the Y-axis,

⇒ T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N

                        \frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N

                    T_1+\sqrt{3}T_2=139.2\times 2

From equation 1, we get

           T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N

                        T_1+3T_1=278.4 \ N

                                4T_1=278.4 \ N

                                  T_1=\frac{278.4}{4}

                                       =69.6 \ N  

6 0
3 years ago
Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
4 years ago
Mrs. Nogaki uses constant force to push a Costco shoping cart down the long store
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Figure B. The second option.
7 0
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