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Ilia_Sergeevich [38]

3 years ago

7

for an object that is speeding up in the positive direction, what does the displacement vs. time graph look like?

2 answers:

Irina18 [472]3 years ago

8 0

Answer:

displacement vs time graph will look like a parabolic shaped graph

Explanation:

As we know that the displacement of an object is given by kinematics equation when object is moving with uniform acceleration

So we know that

$d = v t + \frac{1}{2}at^2$

now here since velocity is increasing in positive X direction as well as the acceleration is also in same direction then the displacement vs time graph must be a straight line graph

so we will have

slava [35]3 years ago

7 0

Answer:

The speeding up is steady, it is a parabola (a=V*t+(at^2)/2), and give it's an equation in connection to time, at that point, it is conceivable to discover the separation recipe by utilizing more substantial amount mathematics(integrals).

Explanation:

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E-mail is usually a poor choice for the distribution of sensitive electronic files because __________. A. Paper copies seem more

iragen [17]

Answer: D.) e-mail is saved on the server that transfers it

Explanation:

3 0

1 year ago

4.7 cm. The tube is sealed at one end and loaded with lead shot to give it a total mass of M = 130 g. The tube floats in water (

babunello [35]

Answer:

Depth is 7.49cm

Explanation:

Given that;

The diameter of tube is (D)=4.7cm

Area of the tube is =πD^2/4

Area is 3.142*4.7*4.7/4=17.35cm2

Mass(M)=130g

Density =1g/cm^3

Recall that Density =mass /volume

Hence volume =130/1=130cm^3

Also note

Volume=area*depth

130=17.35*depth

Depth=130/17.35=7.49cm

6 0

2 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 <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

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

2 years ago

What is the mass of an object that weigh 150 N on earth

aev [14]

450 g =0.9921 lb I hope this helps you out.

8 0

3 years ago

Is the amplitude just 2? Or do I combine all of them and do 6?

oksian1 [2.3K]

Answer:

combine them all

6 m

Explanation:

7 0

2 years ago