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sweet [91]
1 year ago
10

Two objects interact with each other and with no other objects. Initially object A has a speed of 5 m/s and object B has a speed

of 10 m/s. In the course of their motion they return to their initial positions. Then A has a speed of 4 m/s and B has a speed of 7 m/s. We can conclude
Physics
1 answer:
Radda [10]1 year ago
4 0

Answer:

We can conclude that there is a decrease in kinetic energy of the particles due to their elastic collision, since kinetic energy is directly proportional to squared velocity of the particles.

Explanation:

Given:

initial velocity of particle A, Ua = 5m/s

initial velocity of particle B, Ub = 10 m/s

final velocity of particle A, Va = 4m/s

final velocity of particle B, Vb = 7m/s

For particle A:

The final velocity is 1 less than the initial velocity.

For particle B:

The final velocity is 3 less than the initial velocity.

We can conclude that there is a loss in kinetic energy due to elastic collision of the two particles, since kinetic energy is directly proportional to squared velocity of the particles. A decrease in velocity means decrease in kinetic energy.

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For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar
Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

T=\dfrac{30}{2}\ s= 15 \ s

Therefore, total time take to  get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0
1 year ago
As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the
san4es73 [151]

Answer:

conserved

Explanation:

During this process the energy is conserved

3 0
1 year ago
Read 2 more answers
10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
2 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
1 year ago
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
1 year ago
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