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3 years ago

Joe's fast is how many days Long? * 40 50 60

Physics

2 answers:

pickupchik [31]3 years ago

7 0

40?? please explain your question

Bingel [31]3 years ago

5 0

Uh 40 I guess? Lollllll

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Please guys i need help in this

JulijaS [17]

Answer:

10 seconds

Explanation:

We have the equation V = at (speed = acceleration x time)

We want to find the time, so can rearrange to T = V/a (time = speed / acceleration).

From the question, we know V is 5 and a is 0.5.

Now we can substitute that into our equation: 5/0.5 = 10.

So the time is 10 seconds.

Hope this helps! Let me know if you have any questions :)

8 0

3 years ago

if a cat is running at a constant speed of 10km/h for 5 s, what is its average speed and what is its instantaneous speed at 4 s?

iris [78.8K]

Here a cat is running at constant speed which is given as 10 km/h for 5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

$v_{avg} = \frac{distance}{time}$

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

average speed = 10 km/h

instantaneous speed = 10 km/h

8 0

3 years ago

In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the magnitude of the net electric field will be the E1 + E2.

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

3 years ago

How much Nitrogen (N) atoms are in this 3NH4Cl?

mihalych1998 [28]

Answer:

Explanation:

7 0

3 years ago

Explain why is it easy to slip on a floor that is wet

zmey [24]

The water creates less friction between your foot and the ground

6 0

3 years ago

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