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2 years ago

6

What is question 22?

2 answers:

zalisa [80]2 years ago

5 0

The object moves 18 meters in 6 seconds (10 - 4).

18m/6s= 3m/s

Answer "A"

Fynjy0 [20]2 years ago

4 0

the answer is a!! its pretty simple I just read the graph.

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

2 years ago

A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp

Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0

1 year ago

Convert 4 µL to mL<br> Please help meeeee

stiv31 [10]

Answer: the answer would be four thousand

Explanation: hope this helps

3 0

2 years ago

Read 2 more answers

An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle

Katarina [22]

Answer:

The change on the second particle is $2.93\times 10^{-16}\ C$ .

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

$T=\dfrac{2\pi m}{Bq}$

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

$m_s=m_p$

If both particles take the same amount of time to go once around their respective circles. So,

$T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C$

So, the change on the second particle is $2.93\times 10^{-16}\ C$ .

7 0

2 years ago

Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris

telo118 [61]

Answer:

a

$F_A =425.42 \ N$

b

$F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

The diameter of the Ferris wheel is $d = 80 \ ft = \frac{80}{3.281} = 24.383$

The period of the Ferris wheel is $T = 24 \ s$

The mass of the passenger is $m_g = 40 \ kg$

The apparent weight of the passenger at the lowest point is mathematically represented as

$F_A_L = F_c + W$

Where $F_c$ is the centripetal force on the passenger, which is mathematically represented as

$F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = \frac{2* \pi }{T}$

substituting values

$w = \frac{2* 3.142 }{24}$

$w = 0.2618 \ rad/s$

and r is the radius which is evaluated as $r = \frac{d}{2}$

substituting values

$r = \frac{24.383}{2}$

$r = 12.19 \ ft$

So

$F_c = 40 * 12.19* (0.2618)^2$

$F_c = 33.42 \ N$

W is the weight which is mathematically represented as

$W = 40 * 9.8$

$W = 392 \ N$

So

$F_A = 33.42 + 392$

$F_A =425.42 \ N$

The apparent weight of the passenger at the highest point is mathematically represented as

$F_A_H = W- F_c$

substituting values

$F_A_H = 392 - 33.42$

$F_A_H = 358.58 \ N$

4 0

2 years ago