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4 years ago

HELP PLEASE

Physics

2 answers:

Leto [7]4 years ago

8 0

Answer:

2.5m/s^2

Explanation:

We have the first equation of motion here:

V(f) = V(i) + at (equation 1)

Now,

V(f) = Final velocity = 25 m/s

V(i) = Initial velocity = 0 m/s (Since car starts moving from rest, so intial velocity has to be considered as 0 m/s)

a = Acceleration = ? (It has to be determined)

t= Time = 10 seconds

Now by putting values in equation 1

25 = 0 + (a)*(10)

25 = 10*a

a = 25/10

a = 2.5 m/s^2

Additionally, remember that we have three equations of motion

First equation is V(f) = V(i) + at

Second is S = V(i)t + (1/2)a(t^2)

Third is 2as = [V(f)]^2 - [V(i)]^2

Where S = Distance

t = Time

a= Acceleration

V(f) = Final velocity

V(i) = Initial velocity

We always decide to use any of these equations according to the data given, like you can see that I used first equation because it is independent of Distance, since in data, there was no distance given, so i used first equation of motion.

4vir4ik [10]4 years ago

7 0

Recall this kinematic equation:

a = $\frac{Vi+Vf}{Δt}$

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s

Substitute the terms in the equation with the given values and solve for a:

a = $\frac{0+25}{10}$

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A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra

sergey [27]

Answer:

$\frac{di}{dt} = 7.31 \ A/s$

Explanation:

From the question we are told that

The number of turns is $N = 450 \ turns$

The radius is $r = 1.17 \ cm = 0.0117 \ m$

The position from the center consider is x = 3.45 cm = 0.0345 m

The induced emf is $e = 8.20 *10^{-6} \ V/m$

Generally according to Gauss law

$\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A$

Where A is the cross-sectional area of the solenoid which is mathematically represented as

$A = \pi r ^2$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2$

=> $\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}$ ggl;

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

=> $\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}$

=> $\frac{di}{dt} = 7.31 \ A/s$

6 0

4 years ago

What happens to the gravitational force between two objects as the objects move closer together?

zavuch27 [327]

Answer:

The Gravity gets stronger

Explanation:

Newtons laws of motion.

3 0

3 years ago

a parallel plate capacitor has square plates that have edge length equal to 100cm and are separated by 1 mm. It is connected to

Flura [38]

Answer:

the energy is stored in the capacitor is 0.32 μJ

Explanation:

Given;

distance of separation, d = 1 mm = 0.001 m

edge length of the square = 100 cm

potential difference across the plates, V = 12 v

let the side of the square = L

This edge length is also the diagonal of the square which makes a right angle with the side of the square.

Applying Pythagoras theorem;

L² + L² = 100²

2L² = 100²

L² = 100²/2

Note area of a square is L²

A = L² = 100²/2 = 5000 cm²

A (m²) = 5000 cm² x 1m²/(100 cm)²

A = 5000 cm² x 1m²/10000 cm²

A = 0.5 m²

Energy stored in a parallel plate capacitor, E= ¹/₂CV²

C = ε₀A/d

where;

ε₀ is permittivity of free space = 8.85 x 10⁻¹² F/m

d is the distance of separation = 0.001 m

A is the area of the plate

C = ε₀A/d = (8.85 x 10⁻¹²)x0.5 / 0.001

C = 4425 x 10⁻¹² F

E = ¹/₂CV² = ¹/₂ x 4425 x 10⁻¹² x ( 12)²

E = 318600 x 10⁻¹² = 0.32 μJ

Therefore, the energy is stored in the capacitor is 0.32 μJ

8 0

3 years ago

Read 2 more answers

A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is

Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12 m

Let assume the tank have a thickness = dh

The radius of the tank by using the concept of similar triangle is :

$\dfrac{1}{r} = \dfrac{12}{h}$

$r = \dfrac{h}{12}$

The area of the tank = $\mathbf{\pi r^2}$

The area of the tank = $\mathbf{\pi( \dfrac{h}{12})^2}$

The area of the tank = $\mathbf{ \dfrac{\pi}{144}h^2}$

The volume of the tank is = area × thickness

= $\mathbf{ \dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\rho_ g * volume$

where;

$\rho_g$ = density of water ; which is given as 10000 N/m³

So;

Weight of the element = $\mathbf{ 10000 *\dfrac{\pi}{144}h^2 \ dh}$

Weight of the element = $\mathbf{69.44 \ \pi \ h^2 \ dh}$

However; the work required to pump this water = weight × height rise

where the height rise = 12 - h

the work required to pump this water = $\mathbf{69.44 \ \pi \ h^2 \ dh}$ (12 - h)

the work required to pump this water = $\mathbf{69.44 \pi (12h^2-h^3)dh}$

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = $\mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}$

= $\mathbf{ 69.44 \pi[ \frac{12h^3}{3}- \frac{h^4}{4}]^{12}}_0} }$

= $\mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}$

= $\mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}$

= $\mathbf{69.44 \pi*12^4 *\frac{1}{12}}$

= 376966.991 Joules

6 0

3 years ago

How will the behavior of the gases in the piston chamber change if more pressure is applied to the piston?

allsm [11]

Here molecules of gas remains constant. If more pressure is applied to the piston, gaseous molecules will come closer to each other. Hence their volume decreases and density increases. Hope this helps you.

6 0

3 years ago

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