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Sveta_85 [38]
3 years ago
11

A m=7.2 gram object is accelerated at a rate of a=2.9 m/s^2. What force (in millinewtons) does the object experience? No need to

add the unit (already given).
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

20.88 mN

Explanation:

given data:

mass of object = 7.2 gm

acceleration of object = 2.9 m/s2

we know that force is given as

F = ma

where m is mass of object and a is acceleration of moving object.

putting all value to get required force

    = 7.2*10^{-3}\ kg *2.9 m/s2

   = 20.88*10^{-3} N

force in milli newton is

  = 20.88*10^{-3} * 1000 = 20.88 mN

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A certain oscillator satisfies the equation of motion Initially the particle is at
Flura [38]

Answer:

in simple harmonic motion, the formula for finding amplitude is:

a = -ω^2 x

where, a=amplitude, ω= angular frequency, x= Displacement

Explanation:

7 0
3 years ago
A 3.00 kg toy falls from a height of 1.00 m. What will the kinetic energy of the toy be just before the toy hits the ground? (As
yulyashka [42]
Kinetic energy formula: Ek = m · v² / 2
We have: m= 3 kg,  v= ?
h= g·t²/2 ,     t²= 2h/g = 2 m / 9.81 m/s² = 0.2 s² ,  t=√0.2(sqrt)= 0.45 s
v= g· t = 9.81 m/s² · 0.45 s = 4.43 m/s
Finally: Ek=\frac{m v^{2} }{2} = \frac{3 kg * (4.43 m/s)^{2} }{2} = \frac{58.87 J}{2}= 29.43 J
7 0
3 years ago
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by y=40t−16t2. (a) Find
andrew11 [14]

Answer:

(a) (1) -32ft/s

    (2) -25.6ft/s

    (3) -24.8ft/s

    (4) -24.16ft/s

(b) -24 ft/s

Explanation:

First of all, we need the position for every instant mentioned in order to calculate average velocities:

Y(2) = 2 ft

Y(2.5) = 0 ft

Y(2.1) = 13.44 ft

Y(2.05) = 14.76 ft

Y(2.01) = 15.7584 ft

Let's now calculate average velocities:

V_{2-2.5}=\frac{Y(2.5)-Y(2)}{2.5-2}=-32ft/s

V_{2-2.1}=\frac{Y(2.1)-Y(2)}{2.1-2}=-25.6ft/s

V_{2-2.05}=\frac{Y(2.05)-Y(2)}{2.05-2}=-24.8ft/s

V_{2-2.01}=\frac{Y(2.01)-Y(2)}{2.01-2}=-24.16ft/s

Now, for the instantaneous velocity, we derive the expression for y:

V(t) = 40 - 32t    For t=2s   V(2) = -24ft/s

8 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
3 years ago
The upward force exerted on an object falling through air is _____.
stira [4]

(C) Air Resistance

<u>Explanation:</u>

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

6 0
4 years ago
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