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Lubov Fominskaja [6]
3 years ago
12

A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi

th a velocity of 5m_s, find the velocity of 2kg mass​
Physics
1 answer:
Katen [24]3 years ago
7 0

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

P=m*v\\

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

P=4*5\\P=20[kg*m/s]

This same momentum is equal for the other mass, in this way we can find the velocity.

P=m*v\\20=2*v\\v=10[m/s]

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Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe
lutik1710 [3]

Answer:

From the formula of force:

F =  \frac{kAB}{ {r}^{2} }  \\

since AB and k are constants:

F \:  \alpha  \:  \frac{1}{ {r}^{2} }  \\  \\ F =  \frac{x}{ {r}^{2} }

x is a constant of proportionality

• when force is 4N, separation distance is 1

4 =  \frac{x}{1}  \\ x = 4

therefore, equation becomes

F =  \frac{4}{ {r}^{2} }  \\

when r is doubled, r becomes 2. find F:

F =  \frac{4}{ {2}^{2} }  \\  \\ F =  \frac{4}{4}  \\  \\ { \underline{force \: is \: 1N}}

5 0
3 years ago
What's is meant by relative motion?​
saveliy_v [14]

Relative motion means a motion relative to a reference point. We can also say, relative motion means motion referred or observed from a reference point.

For example, Alex is in a train and Ace is at the station, when the train starts moving, for Ace it is moving at a speed of 10 m/s, but for Alex it is moving at 0 m/s, or we can say that it is at rest for Alex, but in motion for Ace. This is relative motion.

4 0
3 years ago
Read 2 more answers
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0
3 years ago
You are watching a wave with a frequency of 512 Hz go past. How many peaks of the wave go past you each minute?
inna [77]

Answer:

61440 peaks

Explanation:

A hertz represents one cycle for every second:

f= Hz=\frac{1}{s}

So if a wave have frequency of 2Hz for example, this means the wave does two cycles per second.

Normally the waves are represented by cosine or sine waves. These kind of waves have two peaks in each cycle, one positive and one negative. With this in mind, let's calculate how many peaks of the wave pass each minute.

A minute has 60 seconds, hence:

60*512=30720\hspace{3}cycles\hspace{3}per\hspace{3}minute

And we know already that every cycle has two peaks, so:

30720*2=61440\hspace{3}peaks\hspace{3}per\hspace{3}minute

4 0
3 years ago
Read 2 more answers
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
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