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KonstantinChe [14]
1 year ago
11

A ball is rolling down a hill. At the initial point of measurement, the ball is traveling at 1.3 m/s. After 15 seconds, the ball

is measured to be traveling at 4.8 m/s.
What is the acceleration of the ball?
Responses

1.3 m/s²
1.3 m/s²
4.8 m/s²
4.8 m/s²
3.5 m/s²
3.5 m/s²
0.23 m/s²
Physics
1 answer:
Rudiy271 year ago
3 0

Answer:

yaaa

Explanation:

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Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.
tresset_1 [31]
We are given with the data that the angle of velocity is above the negative axis. hence, the angle lies in the second quadrant with negative x axis and positive y axis. In this case, the x component is -12 cos 60 equal to -6 m/s while the y-component is 12 sin 60 equal to 6 sqrt of 3 m/s. 
6 0
3 years ago
In the diagram, 91, 92, and q3 are in a straight line. Each of these particles has a charge of -2.35 x 10-6 C. Particles q₁ and
Strike441 [17]

The net force on particle q₃ is  6.2128125 N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/d²

where k = 9 x 10⁹ N.m²/C²

Given is the diagram in which each of the particles has a charge of -2.35 x 10⁻⁶ C. Particles q₁ and q₂ are separated by 0.100 m and particles q₂ and q₃ are separated by 0.100 m.

Force acting on q₃ due to q₁

F₃₁  = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1)²

F₃₁ = 4.97025 N(in right direction)

Force acting on q₃ due to q₂

F₃₂ = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1+0.1)²

F₃₂ = 1.2425 N (in right direction)

Net force acting on particle q₃ is

F₃ = F₃₁ +F₃₂

F₃ = 4.97025 N + 1.2425 N

F₃ = 6.2128125 N

Thus the net force on a charged particle is  6.2128125 N to the right.

Learn more about electrostatic force.

brainly.com/question/9774180

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3 0
2 years ago
A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2
nlexa [21]
Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration a_1 = 2.65~m/s^2 and for a total time of t_1=17~s. The body is initially at rest, so the distance covered is given by
S= \frac{1}{2}a_1t_1^2=382.9~m
Calling v_f and v_i the final and initial velocity, and since the v_i=0~m/s because the body starts from rest, we can use
a= \frac{v_f-v_i}{t}
to find the final velocity after this first leg:
v_{f}=v_i+a_1t_1=45~m/s
And the average velocity in this first leg is
v_1= \frac{v_f+v_i}{2}=22.5~m/s

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: v_2=45~m/s. This is also the average velocity of the second leg. 
The total time of this second leg is t_2=1.60~min = 96~s. The distance covered is given by
S_2=v_2t_2=45~m/s \cdot 96~s=4320~m

3) Uniformly decelerated motion, with constant deceleration of a_3=-9.39~m/s^2 and for a total time of t_3=4.8~s. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. v_i=45~m/s. Therefore, the distance covered in this leg is given by
S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m
The final velocity in this leg is given by
v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
v_3 =  \frac{1}{2}(v_f+v_i)=  \frac{1}{2}(-0.07~m/s+45~m/s)=  22.5~m/s

4) Finally, the total distance covered in the motion is
S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s
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Explanation:

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Answer:

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