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3 years ago

HELP Please Hurry

Physics

2 answers:

nlexa [21]3 years ago

6 0

The correct answers are:

A plane landing on an aircraft carrier.

Rain sticking to a window.

Two train cars coupling together.

2. The total momentum is zero.

3. 3.6 (kg*m/s) (Option C)

Explanations:

1) In simple terms (not a textbook definition), perfectly inelastic collision is a collision in which two bodies stick together (or becomes one) after a collision. Now let us have a look at the options:

A baseball bouncing off a bat: After collision, the ball and the bat do not stick to one another; therefore, it is NOT a perfectly inelastic collision.

Bumper cars bumping off of each other: After collision, cars bump off of each other, making their collision elastic, not perfectly inelastic collision.

A cue ball hitting an eight ball and stopping: After collision, although the cue ball stops, but all of its momentum will be transferred to the eight ball; and eight ball will start moving, where is cue ball stops. Not a perfectly inelastic collision.

A plane landing on an aircraft carrier: After plan lands, the plane and the aircraft carrier will be incontent with each other; making their collision perfectly inelastic.

Rain sticking to a window: Rain drop sticking to a window means both stick together after a collision, making it a perfectly inelastic collision.

Two train cars coupling together: Again both cars are sticking together, making it a perfectly inelastic collision.

Hence the correct answers are:

A plane landing on an aircraft carrier.

Rain sticking to a window.

Two train cars coupling together

7) Always remember that in a closed system, the total momentum is conserved, meaning:

Total initial momentum = Total final momentum

Initially, bodies are at rest, the total initial momentum (mv) is zero (since (m+2m)*0 = 0; as v = 0). As it is the closed system, the total final momentum will be equal to the total initial momentum. As the total initial momentum is zero, the total final momentum will also be zero.

Hence the correct answer is: The total momentum is zero.

3) As you can see in the table, the initial momentum of each and every entry is equal to the final momentum. In the case of X, as the final momentum is 3.6 kg*m/s, the initial momentum will be same as the final momentum (by considering the pattern in the table); therefore X (the initial momentum) will be 3.6 kg*m/s.

Hence the correct answer is: X = 3.6 (Option C)

Mandarinka [93]3 years ago

3 0

Answer:

A plane landing on an aircraft carrier.

Rain sticking to a window.

Two train cars coupling together.

Explanation:

In perfectly inelastic collision when two objects collide each other then after collision they stick with each other as in this type of collision there is no separation after collision.

so here we have to choose such cases in the given all options

A plane landing on an aircraft carrier:

After landing the aircraft is in always contact with the ground so it is satisfying the condition of inelastic collision.

Rain sticking to a window:

Rain drops sticking to the window after it collide with it so it is the condition of perfect inelastic collision.

Two train cars coupling together:

Two cars collide and stick with each other so this is the condition of perfect inelastic collision.

Answer:

The total momentum is zero.

Explanation:

As we know that it is a type of perfect inelastic collision so here we can use momentum conservation as there is no external force on it.

Since we know that if net external force on a system is zero then momentum of the system is always conserved.

So in this case both the masses are at rest initially so initial momentum is zero here so at all instants of time final momentum is also zero.

so total momentum is zero

Answer:

Option C 3.6 (kg*m/s)

Explanation:

Since here in all cases we can see that it is following the condition of momentum conservation

Here by momentum conservation we can say that

initial momentum = final momentum

so from given table we know that

final momentum = 3.6 kg m/s

so initial momentum must be same

so value of x = 3.6 kg m/s

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4. What quantity of heat is required to raise the temperature of 100

kupik [55]

Answer:

Q = 836.4 Joules.

Explanation:

Given the following data;

Mass = 100 grams

Initial temperature = 25°C

Final temperature = 45°C

We know that the specific heat capacity of water is equal to 4.182 J/g°C.

To find the quantity of heat;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 45 - 25

dt = 20°C

Substituting the values into the equation, we have;

$Q = 100*4.182*20$

Q = 836.4 Joules.

5 0

3 years ago

Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to report the mass truthfully). (a) W

bija089 [108]

Answer:

1411.8 N/m

Explanation:

From Hooke's law;

F= Ke

Where

F= force on the spring

K= force constant

e = extension

But e= 8.50 × 10^-2m

F= weight = 12.0 kg × 10 = 120 N

K = F/e = 120/8.50 × 10^-2

K= 1411.8 N/m

7 0

3 years ago

g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

3 years ago

A 48-kg woman pushes a 12-kg grocery cart with a force of 24 N. What is the magnitude of the force that the grocery cart exerts

Nata [24]

The correct answer is B. 24 N.

To figure out the exact force the cart exerts on the woman, we have to use Newton's third law of motion. Newton's third law of motion states that every action has an equal and opposite reaction. This means that if object 1 exerts a force on object 2 , object 2 will exert an equal but opposite force on object 1.

This law allows us to ignore the masses of the chart and the women and focus on the pairs of forces the woman and the chart apply to each other. Since the woman exerts a force of 24 N on the chart, the chart will exert a force of 24 N on the woman.

The correct answer is B. 24 N.

4 0

3 years ago

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0

3 years ago