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DaniilM [7]
3 years ago
13

A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?

Physics
2 answers:
loris [4]3 years ago
6 0
The car's acceleration is      (6 m/s) / (12 sec)  =  0.5 m/s² .

           Force = (mass) x (acceleration)

                     =  (2,300 kg) x (0.5 m/s²)

                     =    1,150 Newtons
pav-90 [236]3 years ago
4 0
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs
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Suppose you have a 75-watt light bulb that you leave turned on for two minutes. How much energy
victus00 [196]

Answer:

9000 J

Explanation:

Convert minutes to seconds.

2 min = 120 s

Power = energy / time

75 W = E / 120 s

E = 9000 J

5 0
2 years ago
Means/end analysis involves:_________
polet [3.4K]

a. a gradual approximation to the final solution

Explanation:

Means/end analysis is a process that involves creation of an end goal to enable the identified means to apply.

In this techniques sub-goals are formed to eliminate the challenges faced in application of a selected operator.

It starts by identifying a predetermined goal which is followed by actions that will led to the goal.

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4 0
3 years ago
A pencil has a density of 0.875 g/ml. It has a volume of 4.0 ml. Find the mass
loris [4]

Answer:

To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.

mass=density*volume

mass=0.875\frac{g}{ml}*4.0\ ml

mass=3.5\ g

Therefore, our final answer is that our pencil weight 3.5 grams

<u><em>Hope this helps!  Let me know if you have any questions</em></u>

3 0
2 years ago
A 1200 kg car accelerates from 13m/s to 17m/s. find the change in momentum of the car?
lisabon 2012 [21]

Answer:

4800

Explanation:

The change in velocity of the car is 17-13=4m/s. Since the change in momentum is the mass multiplied by the change in velocity, the answer is 4*1200=4800. Hope this helps!

6 0
3 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
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