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3 years ago

Help me please Am I correct???

Physics

1 answer:

trasher [3.6K]3 years ago

5 0

Answer:

They both experienced the same force as they weigh the same amount

Newtonian physics

Explanation:

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What is the electric force acting between two charges of 0.0072 C and -0.0060 C that are 0.0040 m apart?

telo118 [61]

Answer:
F = -2.4 * 10¹⁰ N

Explanation:
To get the electric force between two charges, we would use Coulomb's law which states that:
F = $\frac{k * q1 * q2}{d^{2} }$

where:
F is the force between the two charges
k is Coulomb's constant = 9 * 10⁹ Nm²/C²
q1 is the first charge = 0.0072 C
q2 is the second charge = -0.006 C
d is the distance between the two charges = 0.004 m

Substitute with the givens in the above formula to get the force between the two charges as follows:
F = $\frac{9*10^9*0.0072*-0.006}{(0.004)^2}$

F = -2.4 * 10¹⁰ N

Hope this helps :)

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3 years ago

What is the acceleration at 3 seconds?<br><br> Please explain your answer. Thank you :]

statuscvo [17]

Answer:

$-3 m/s^{2}$

Explanation:

Acceleration on a VT graph is the slope of the line at the given point. We can find the slope at 3 with Δy/Δx. This gives us (4-2)/(3-(-3)) which works out to be -3m/s^2

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3 years ago

Which term means that humans are combining two different organism's DNA to create a new organism. Hint: Oink, Oink Question 3 op

cestrela7 [59]

Answer:

breeding

Explanation:

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3 years ago

Dormancy helps plants survive freezing temperatures and lack of..

quester [9]

I’m not 100% but I was thinking A

6 0

3 years ago

A race car starting from rest accelerates uniformly at a rate of 4.90 meters per squared. What is the cars speed after it has tr

shusha [124]

Ok, we need to find a relation for the speed as it relates to the acceleration. This is given by the integral of acceleration:

$v= \int\limits^{t}_{0} {a} \, dt' =at$

Where we have the initial velocity is 0m/s and a will be 4.90m/s².

But we see there is an issue now... We know the velocity as a function of time, but we don't know how long the car has been accelerating! We need to calculate this time by now finding the position function as a function of time. This way we can solve for the time, t, that it takes to go 200m accelerating this way and then substitute that time into our velocity equation and get the velocity.
Position is just the integral of velocity:

$s= \int\limits^{}_{} {at} \, dt = \frac{1}{2}at^2$

Where the initial velocity and initial position are both zero.

Now we set this position function equal to 200m and find the time, t, it took to get there

$\frac{1}{2}(a \frac{m}{s^2} )t^2=200m \\ \\ \frac{1}{2}4.90 \frac{m}{s^2} t^2=200m \\ \\ t^2= \frac{400m}{4.90 \frac{m}{s^2}}=81.63s^2 \\ \\ t= \sqrt{81.63s^2 } =9.04s$

Now let's put t=9.04s into our velocity equation:

$v =at=4.9\frac{m}{s^2} \times 9.04s=44.3 \frac{m}{s}$

8 0

3 years ago

Help me please Am I correct???​

Help me please Am I correct???