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2 years ago

Why is an important to come up with a plan for how are you to respond to negative peer pressure before the situation happens

Physics

1 answer:

natima [27]2 years ago

4 0

Answer:

C ( is hard to refuse in the situation if you have not decided how you would handle it ahead of time

B: because it is very likely you will face these situation in the future)

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A push of magnitude P acts on a box of weight W as shown in the figure. The push is directed at an angle θ below the horizontal,

Kipish [7]

Answer:

its on wheels and they are supposed to make it eas

Explanation:

3 0

2 years ago

According to the definition of mechanical work, pushing on a rock accomplishes no work unless there is

LekaFEV [45]

There must be movement in the same direction as the force put on the object. Hope this helps!

4 0

3 years ago

Read 2 more answers

A 1000-kg car moving at 10 m/s brakes to a stop in 5 s. the average braking force is

Marina86 [1]

To calcculate the braking force of the car moving, we use Newton's second law of motion which relates the acceleration and the force of an object moving. The force of an object moving is directly proportional to its acceleration and the proportionality constant is the mass of the object. It is expressed as:

Force = ma

Acceleration is the rate of change of the velocity of a moving object. We calculate acceleration from the velocity and the time given above.

a = (10 m/s) / 5 s = 2 m/s^2

So,
Force = ma
Force = 1000 kg ( 2 m/s^2 )
Force = 2000 kg m/s^2 or 2000 N

4 0

3 years ago

Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m

Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.