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Crank
3 years ago
9

a 6-kg and a 4-kg ball are acted on by forces of equal size. if the large ball accelerates at 2 m/s2 what acceleration will the

small ball undergo
Physics
2 answers:
Dennis_Churaev [7]3 years ago
5 0
To answer this question, we need to find the force acting on the two objects.

f=ma
f=(6)(2)
f=12 newtons

So, now we can determine the acceleration of the second object using the same formula!

f=ma
12=(4)a
3=a

So the acceleration of the smaller object is 3 meters per second squared!
Anna007 [38]3 years ago
3 0

It would be 30 meters.

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In a laboratory experiment, a computer determines that the time for a falling steel ball to travel the final 0.80 m before hitti
Roman55 [17]

Answer:

Vf = 9.622 m/s

Explanation:

Distance covered S=0.80m, Time t=0.087s and

g=9.81 m/s²  

Required: Final velocity Vf when it hit the floor

Sol:

We have first find Vi so

Vi=Vit + 0.5 a t2

Vi =8.768m/s

Now Vf= Vi + at

Vf = 9.622 m/s

8 0
4 years ago
If a 4kg Bird is pushed by the window with a force of 60 N how fast is the bird accelerate?
LiRa [457]
  • Force=60N
  • Mass=4kg

\\ \ast\sf\hookrightarrow Force=Mass\times Acceleration

\\ \ast\sf\hookrightarrow Acceleration=\dfrac{Force}{Mass}

\\ \ast\sf\hookrightarrow Acceleration=\dfrac{60}{4}

\\ \ast\sf\hookrightarrow Acceleration=15m/s^2

3 0
3 years ago
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic
olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma              

a = \frac{F}{m}      ...... (1)

acceleration (a) = \frac{dv}{dt}   ......(2)

substituting (2) into (1)

Hence, F = \frac{mdv}{dt}

\frac{dv}{dt} = \frac{F}{m}

dv = \frac{F}{m} dt

dv = \frac{1}{m}Fdt

Integrating both sides

\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt     ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt

v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}

v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|

v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |

v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |

v = \frac{1}{5} | 250 - 50 + 15 |

v = \frac{215}{5}

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0
3 years ago
6.
Vikentia [17]

Answer:

Explanation:

Givens

vi = 0

a = 9.81

d = 4.50 m

vf = ?

Formula

vf^2 = vi^2 + 2 * a * d

Solution

Substitute the knowns into the formula

vf^2 =0 +  2 * 9.81 * 4.50

vf^2 = 88.29                          Take the square root of both sides.

sqrt(vf^2) = sqrt(88.29)    

vf = 9.40 m/s

3 0
3 years ago
A watt is a unit of<br><br>light<br>power<br>force<br>motion<br>energy
Crazy boy [7]

The answer would be Power

3 0
3 years ago
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