Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

Read 2 more answers

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

stellarik [79]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

8 0

3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

6 0

3 years ago

A train travels 76 kilometers kilometers in 2 hours and then 54 kilometers in 5 hours . What is tuts average speed?

Elena-2011 [213]

Total distance = 76+54 = 130km
total time = 2+5 = 7hrs
Av. speed = 130/7 = 18.571km/hr = 18.6 km/h ( 3 sig fig)

7 0

3 years ago

5.3 two 30 kg children in a 20 kg cart are stationary at the top of a hill. They start rolling down the 80 m tall hill and they

Makovka662 [10]

Answer:

60008.4 J

Explanation:

The mass of each kid = 30 kg

mass of the cart = 20 kg

The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s

The height of the hill = 80 m

The potential energy of the boys at the top of the hill = mgh

where

m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg

g is the acceleration due to gravity = 9.81 m/s^2

h is their height above the ground = 80 m (on the top of the hill)

substituting, we have

potential energy PE = 80 x 9.81 x 80 = 62784 J

At an instance at the bottom of the hill

their kinetic energy = $\frac{1}{2} mv^2$

where

v is their velocity = 8.33 m/s

m is their total mass = 80 kg

substituting, we have

kinetic energy KE = $\frac{1}{2}*80*8.33^2$ = 2775.6 J

Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill

work done by friction = PE - KE = 62784 - 2775.6 = 60008.4 J

5 0

3 years ago