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vredina [299]
3 years ago
14

I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second

s. The acceleration is a constant. How much is the acceleration?
Physics
2 answers:
kozerog [31]3 years ago
8 0

Answer:

-1.4 m/s²

Explanation:

The initial velocity is 7 m/s.  When the ball rolls back, its velocity is -7 m/s.

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (-7 m/s − 7 m/s) / 10 s

a = -1.4 m/s²

Degger [83]3 years ago
3 0

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff ro
dexar [7]

Required value of initial speed of the bullet be ( 4M/m)√(gL).

Given parameters:

Mass of the bullet =m.

Mass of the bob of the pendulum = M.

speed of the bullet before collision = v

Speed of the bullet after collision = v/2.

Length of the pendulum stiff rod = L.

Let speed transmitted to the pendulum be u.

Using principle of conservation of momentum:

mv = Mu + mv/2

⇒ Mu = mv/2

⇒ u = (m/M)v/2

We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]

To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:

u = √(4gL)

⇒  (m/M)v/2 = √(4gL)

⇒ v =( 4M/m)√(gL).

Hence, minimum required  speed of the bullet be ( 4M/m)√(gL).

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brainly.com/question/28224010

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7 0
1 year ago
Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0
2 years ago
Read 2 more answers
What are the three longest wavelengths for standing waves on a 264- cm -long string that is fixed at both ends
miv72 [106K]

The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

  1. 5.2 meters.
  2. 2.6 meters.
  3. 1.7meters.

Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

where,

  • λ is the wavelength
  • L is the length of the string

For n = 1,

  • λ = 2×2.6/1

= 5.2 meters

For n = 2,

  • λ = 2×2.6/2

= 2.6 meters

For n = 3,

  • λ = 2×2.6/3

= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

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7 0
2 years ago
In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro
ANTONII [103]
Is there a picture?
6 0
3 years ago
[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the
ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley (r_p) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw

t = \frac{4r^2mw}{Tr_P}

Substituting values:

t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s

7 0
3 years ago
Read 2 more answers
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