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3 years ago

I need help with this too. (im not good at science or math)

Physics

2 answers:

dedylja [7]3 years ago

7 0

I believe 1 m/s.....

mr_godi [17]3 years ago

6 0

Answer:

i belive 1 m/s

Explanation:

dividing displacement from time it should be 1 cuz 5/5 is 1

please tell me if right!

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The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy correspo

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Answer:

Explanation:

2.1 x 10^2 - 20J

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2 years ago

I NEED ANSWERS ASAP!!!!

BartSMP [9]

Answer: Force (F=m*a)

Explanation:

7 0

3 years ago

Read 2 more answers

Kim travelled at a speed if 18km/h for two hours. What was the distance covered?

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Answer:

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4 years ago

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial:

$0.8\cdot t^{2}-56.92\cdot t +3.872 = 0$ (3)

$t_{1} \approx 71.082\,s$ , $t_{2}\approx 0.068$

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

3 0

3 years ago

Question 4 (1 point)

djverab [1.8K]

Answer:

directly proportional.

Explanation:

Coulomb's has the formular :

F = kQ₁Q₂ / d²

from the formular F = force

also from the formular Q = charge

hence, from the equation above, force and charge are directly related.so if the charge doubles, the force exerted by it on another charge also doubles because they are directly proportional.

5 0

2 years ago