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coldgirl
3 months ago
8

A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho

rizontal distance 3d . Which expression represents the change in the kinetic energy of the object?
Physics
1 answer:
Softa [3K]3 months ago
5 0

The alteration in kinetic energy is \Delta K = 3Fd

Clarification:

According to the work-energy principle, the task performed on an object corresponds to the alteration in its kinetic energy. In mathematical terms:

W=K_f -K_i= \Delta K

where:

W signifies the work performed on the object

K_f denotes the kinetic energy at the end

K_i indicates the kinetic energy at the start

Furthermore, when the force is exerted in line with the object’s motion, the work done is expressed as:

W=F\Delta x

Here,

F represents the force’s magnitude

\Delta x denotes the object’s displacement

In this scenario, the force impacting the object is

F

While the distance moved is the horizontal length traveled, hence

\Delta x = 3d

Consequently, the work accomplished is

W=(F)(3d)=3Fd

Thus, the alteration in kinetic energy amounts to

\Delta K = 3Fd

Learn more about work and kinetic energy:

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Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.
serg [3582]

Answer:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

Consequently, we find that:

V_A = \frac{1}{4} V_B

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

L_A = L_B =L both wires share the same length

both wires carry an identical currentI_A = I_B =I

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

n_A = n_B =n

We also know that r_A = 2 r_B where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B

A_B = \pi r^2_B

Thus, we conclude that

A_A = 4 A_B

Now we are aware that the drift velocity of an electron in a wire can be described by:

v_d = \frac{I}{neA}

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}

V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}

So we observe that:

V_A = \frac{1}{4} V_B

Thus, the most fitting answer is:

a. vA = vB/4

6 0
2 months ago
Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house
kicyunya [3294]
1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.
6 0
3 months ago
You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
serg [3582]
The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.
8 0
3 months ago
A person shooting at a target from a distance of 450 metres finds that the sound of the bullet hitting the target comes 1 / 2 se
Sav [3153]

Answer:

1350 m/s

Clarification:

Bullet speed

The bullet travels 450 m

Sound travels a distance of 450 m

Using the equation S= V × t

==> t= S/V

Thus, the time for the bullet t1=450/vb

and the sound's travel time t2=450/vs

Given that there's a 1/2 sec interval from when the shot is fired to the moment the shooter hears the sound

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec  ---- (1)

At a distance 'x' from both the gun and target, it takes 3 seconds for a person to hear the bullet sound from firing to impact.

Firing sound duration

n

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

During this time period, the bullet covers 450 m and the sound of impact travels a distance 'x'

The time taken for sound = 450/vb

The time it takes sound to travel distance 'x'= x/vs

therefore let T2= 450/vb + x/vs

However, all this occurs within 3 seconds, i.e., T = 3 sec

because firing takes place before hitting the target, implying the strike sound is heard in time T = T2-T1= 450/vb + x/vs -x/vs

Making T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From equations 1 and 2

applying the same principle, in 3 seconds the observer sees the bullet travel 450 m and perceives the sound


3 0
2 months ago
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