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Semenov
1 month ago
7

To avoid an accident, a driver steps on the brakes to stop a 1000-kg car traveling at 65km/h. if the braking distance is 35 m, h

ow much force is necessary to stop the car?
Physics
1 answer:
Maru [3.3K]1 month ago
6 0
Initially, we need to calculate the acceleration required for the car to halt from its initial speed based on the distance traveled. This can be done using the formula,
                                             2ad = Vf² - Vi²
where a represents acceleration, d is distance, and Vf and Vi are the final and initial speeds respectively. Plugging in the known quantities,
           2(a)(35 m) = (0 m/s)² - ((65 km/h) x (1000 m/ 1 km) x (1 hr / 3600 s))²
The resulting acceleration is −4.66 m/s².
To calculate the force required to stop the car, we multiply the mass by the acceleration. This calculation yields -4,660 N, and we take the absolute value, which is 4,660 N. 
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At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w
serg [3582]

Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

               E = \frac{\sigma}{2 \epsilon_{o}}

where,   \sigma is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

             F = eE

where,    e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

                       F = ma

                 a = \frac{eE}{m}

                    = \frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}

                     = \frac{e \sigma}{2m \epsilon_{o}}

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

              v_{f} = v_{i} + at

                     = (0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t

                     = \frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}

                     = 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

                v' = \sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}

                    = \sqrt{921600 + 467820.43}

                    = \sqrt{1389420.43}

                    = 1178.73 m/s

Consequently, we conclude that the proton's speed is 1178.73 m/s.

3 0
17 days ago
If the position of an object is zero at one instant, what is true about the velocity of that object?
ValentinkaMS [3465]
If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.
6 0
1 month ago
Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
Sav [3153]
Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.
4 0
15 days ago
Alicia intends to swim to a point straight across a 100 m wide river with a current that flows at 1.2 m/s. She can swim 2.5 m/s
Sav [3153]

Answer:

θ = 61.3°

Alicia must swim at an angle of 61.3°

Explanation:

Parameters given include:

Width of the river = 100 m

Alicia's speed in still water = 2.5 m/s

Speed of river's current = 1.2 m/s

The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.

Her swimming speed aimed against the current must offset the current's velocity;

2.5cosθ - 1.2 = 0

2.5cosθ = 1.2

cosθ = 1.2/2.5

θ = cosinverse(1.2/2.5)

θ = 61.3°

4 0
1 month ago
A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o
Ostrovityanka [3204]

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

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Bottom

    Em_{f} = K = ½ m v²

    Emo =Em_{f}

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

  v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

     F = ma

     a = v² / r

For the turning radius, the cable length is r = L.

    F = m 2g (1 - cos θ)

Now, let's find the result.

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N

   

7 0
28 days ago
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