Ask question

Ask question

All categories

1 year ago

10

A moving bomb explodes into two pieces. A 2 kg fragment moves to the right at 15 m/s and the other fragment of mass 10 kg moves

to the left at 6 m/s. What was the initial speed

1 answer:

Alla [95]1 year ago

8 0

The initial velocity is obtained as 2.5 m/s.

<h3>What is the initial speed?</h3>

To obtain the initial speed, we have to apply the law of conservation of linear momentum which states that momentum before collision is equal to momentum after collision.

Now, the two fragments initially had one velocity before they were split in two directions opposite each other;

Applying the principle of conservation of linear momentum to this problem;

(2 + 10)v = (6 * 10) - (2 * 15) (they moved in opposite directions)

12v =60 - 30

12v = 30

v = 30/12

v = 2.5 m/s

Learn more about momentum:brainly.com/question/24030570

#SPJ1

You might be interested in

A ship was traveling across the Atlantic Ocean. It traveled 2,000 kilometers at a rate of 40 kilometers per hour. How long did i

8_murik_8 [283]

The correct answer is D

6 0

3 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

3 0

3 years ago

Read 2 more answers

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

3 years ago

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

3 years ago