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2 years ago

100 POINTS!! PLEASE DONT ANSWER UNLESS YOU KNOW BOTH ANSWERS PLEASE.

Physics

2 answers:

JulijaS [17]2 years ago

6 0

Answer:

1. is B and 2. is A

Explanation:

Colt1911 [192]2 years ago

6 0

Answer:

1. Displacement

2. 5 m/s 400 ÷ 80 = 5

Explanation:

You might be interested in

How to change 0.000 007 8 s to us?

Greeley [361]

The letter u before any unit means micro. Micro is a unit prefix used in the metric system which actually means a factor of 10^-6 (one millionth).

Now, 10^-6 is equivalent to a movement of 6 decimal places from left to right. So to write the number 0.0000078 s into a number which has a unit of us, therefore we move the decimal place by 6 to the right, giving us:

7.8 us

Another way to do this is to multiply the number by 1 million (1,000,000), so:

0.0000078 * 1,000,000

Using the calculator to solve, we get the value:

7.8 us

8 0

2 years ago

CAN SOMEONE PLEASE ANSWER NUMBER 8 AND 9 FOR ME WITH AN EXPLANATION HURRY I HAVE LIMITED TIME PLEASE!!!

Aneli [31]

Answer:

D because the molecules become solid so dont move as much 9 A because pressure would build up in the tank

Explanation:

6 0

2 years ago

Read 2 more answers

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same dire

Dahasolnce [82]

Answer:

v₁ =0.19 m/s and v₂ = 0.18 m/s

Explanation:

By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

Momentum:

Before (b) = After (a)

$m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}$

$26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ $7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}$ (1)

Energy:

Before (b) = After (a)

$\frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}$

$26.5 kg*(0.205 m/s)^{2} + 12.5 kg*(0.150 m/s)^{2} = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}$ (2)

From equation (1) we have:

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}$ (3)

Now, by entering equation (3) into (2) we have:

$1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}$ (4)

By solving equation (4) for $v_{1a}$ , we will have two values for

$v_{1a} = 0.16$

$v_{1a} = 0.21$

We will take the average of both values:

$v_{1a} = 0.19 m/s$

Now, by introducing this value into equation (3) we can find $v_{2a}$ :

$v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}$

$v_{2a} = 0.18 m/s$

Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

I hope it helps you!

6 0

2 years ago

Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive f

kumpel [21]

Answer:

$r_{final}=r_{i}/4$

Explanation:

We know force between 2 charges is given by

$F=\frac{1}{4\pi \varepsilon }\frac{q_{1}q_{2}}{r^{2}}$

It is given initial force is attractive of magnitude $F_{i}$

Let $r_{i}$ be the initial separation

Thus

$F_{i}=\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$

Now

Let $r_{f}$ be the final separation

Thus final force becomes

$\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

It is given $F_{f}$ = $16F_{i}$ ...................(i)

Using values of $F_{f}$ and $F_{i}$ in equation i we have

$16\times \frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{i}^{2}}$ = $\frac{1}{4\pi \varepsilon }\frac{q_{a}q_{b}}{r_{f}^{2}}$

thus

$thus\\\\(\frac{r_{i}}{r_{f}})^{2}=16\\\\\frac{r_{i}}{r_{f}}=4\\\\\therefore r_{f}=r_{i}/4$

Thus the final separation is one-fourth of the initial separation

7 0

2 years ago

A 100 W engine runs for 1 minute. How much work does it do?

Svet_ta [14]

d 6000 j because every second multiplied by that would equal out to be 6000

4 0

3 years ago

Read 2 more answers