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BaLLatris [955]
3 years ago
11

At a certain time a particle had a speed of 17 m/s in the positive x direction, and 3.0 s later its speed was 28 m/s in the oppo

site direction. What is the average acceleration of the particle during this 3.0 s interval
Physics
1 answer:
dangina [55]3 years ago
5 0

Answer:

i wanna say 11 forgive me if wrong.

Explanation:

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When catching a baseball, the ball applies a force of 39.6 N to a  catcher's glove.  If the work done on the catchers glove is 4
cricket20 [7]

Answer:

d = 1.19 m

Explanation:

Given that,

The force applied by the ball, F = 39.6 N

The work done on the catchers glove is 47.5 J

We need to find the distance traveled by the ball. We know that,

Work done, W = Fd

Where

d is the distance traveled

d=\dfrac{W}{F}\\\\d=\dfrac{47.5 }{39.6 }\\\\d=1.19\ m

So, it will cover 1.19 m.

4 0
3 years ago
Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0
3 years ago
A racquet ball with mass m = 0.221 kg is moving toward the wall at v = 13.9 m/s and at an angle of θ = 25° with respect to the h
Salsk061 [2.6K]

Answer:

1) 3.07kgm/s

2) 5.56kgm/s

3) 76.16N

4) 4.33kgm/s

5) 0.57s

6) -8.66J

Explanation:

Given

m = 0.221kg

v = 13.9m/s

θ = 25°

t = 0.073s

1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,

P(i) = mv(i)

P(i) = 0.221 * 13.9

P(i) = 3.07kgm/s

2) Magnitude of the change in momentum of the ball,

P(i,x) = P(i) cos θ

P(i,x) = 3.07 * cos25

P(i,x) = 3.07 * 0.9063

P(i,x) = 2.78

ΔP = 2P(i,x)

ΔP = 2 * 2.78 = 5.56kgm/s

3) magnitude of the average force exerted by the wall,

F(ave) = ΔP/Δt

F(ave) = 5.56/0.073

F(ave) = 76.16N

4) ΔP(z) = mv(f) - mv(i)

ΔP(z) = 0.221*-7.8 - 0.221*11.8

ΔP(z) = -1.72 - 2.61

ΔP(z) = 4.33kgm/s

5) F(ave) = ΔP/Δt

Δt = ΔP/F(ave)

Δt = 4.33 / 76.16

Δt = 0.57s

6) KE(i) = 0.5mv(i)²

KE(f) = 0.5mv(f)²

ΔKE = 0.5m[v(f)² - v(i)²]

ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]

ΔKE = 0.1105 ( 60.84 - 139.24 )

ΔKE = 0.1105 * -78.4

ΔKE = -8.66J

6 0
3 years ago
A peacock is flying around and its velocity v as a function of time
anyanavicka [17]
The answer is 3 m. This is the area under the graph from t=2 to t=3, using the trapezium rule. 1/2 (2+4) * 1
3 0
3 years ago
Read 2 more answers
The KEOM tower across the street from West is approx. 155 meters high
blondinia [14]

You told your insane friend that it will take the book 5.6 s to get to the ground.

From the question given above, the following data were obtained:

Height (H) = 155 m

<h3>Time (t) =? </h3>

<u>NOTE</u>: Acceleration due to gravity (g) = 10 m/s²

The time taken for the book to get to ground can be obtained as follow:

<h3>H = ½gt²</h3>

155 = ½ × 10 × t²

155 = 5 × t²

<h3>Divide both side by 5</h3>

t^{2}  = \frac{155}{5}\\\\

t² = 31

<h3>Take the square root of both side </h3>

t = \sqrt{31}

<h3>t = 5.6 s</h3>

Thus, the time taken for the book to get to the ground is 5.6 s

Hence, you told your insane friend that it will take the book 5.6 s to get to the ground.

Learn more: brainly.com/question/24903556

7 0
2 years ago
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