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3 years ago

Help me please, help me

Physics

2 answers:

Elenna [48]3 years ago

6 0

Yea what they saidekbekenekef

kolbaska11 [484]3 years ago

4 0

for the question w the coffee cup and the dishwasher: 19,536.72 J

for the question w the 3 pots of boiling water: They all have the same temperature

for the question w the copper: 103.84 Celsius

(Not sure what the other 2 pictures are about but good luck)

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Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$

Now, we calculate P₁ and P₂.

$P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns$

Therefore, the net initial momentum of the system is given as:

$P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns$

The final momentum of the system is given as:

$P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f$

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

$P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s$

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0

3 years ago

(1 point) A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m

sasho [114]

Answer:

The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)

Explanation:

we have to:

m = mass = 4 kg

k = spring constant = 577 N/m

c = damping constant = 4 N*s/m

The differential equation of motion is equal to:

m(d^2x/dt^2) + c(dx/dt) + k*x = 0

Replacing values:

4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0

Thus, we have:

4*x^2 + 4*x + 577 = 0

we will use the quadratic equation to solve the expression:

x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2) ± 12i

The solution is equal to:

x = e^(1/2) * (c1*sin12t + c2*cos12t)

x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)

We have the follow:

x(0) = 5

e^0(0*c1 + c2) = 5

c2 = 5

x´(0) = 7

(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7

(-1/2)*(5) + 12*c1 = 7

Clearing c1:

c1 = 0.792

The function is equal to:

x = e^(-t/2) * (0.792*sin12t + 5cos12t)

8 0

3 years ago

An isotope releases radiation in order to ____________ its nucleus.

makvit [3.9K]

Answer:

stabilize

I think this should be the answer...

4 0

3 years ago

Using carson's rule, what is the required bw to transmit an fm signal if the maximum deviation is 60 khz and intelligence freque

VashaNatasha [74]

Carson's Rule says:

FM occupied bandwidth =

(2) · (Peak deviation + Highest modulating frequency)

FM bw = (2) · (60 kHz + 15 kHz)

FM bw = (2) · (75 kHz)

FM bw = 150 kHz

(I used to eat this stuff for lunch, but it's been almost 40 years. Thanks for taking me back. Those were the good old days.)

5 0

3 years ago

A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by

Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0

3 years ago