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GuDViN [60]
3 years ago
12

A ball has a mass of 2 kg and is thrown with a force of 8 Newtons for .35 seconds. What is the ball's change in

Physics
1 answer:
Anna [14]3 years ago
6 0

Answer:

1.4s

Explanation:

Given parameters:

Mass of ball  = 2kg

Force  = 8N

Time  = 0.35s

Unknown:

Change in velocity  = ?

Solution:

To solve this problem, we use the expression obtained from Newton's second law of motion which is shown below:

      Ft  = m(v  - u)

 So;

         Ft  = m Δv

F is the force

t is the time

m is the mass

Δv is the change in velocity

             8 x 0.35  = 2 x Δv

                  Δv  = 1.4s

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Read 2 more answers
To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0
3 years ago
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