Given an objects mass and observing its acceleration what can be found in a direct application of Newton’s second law of motion?

A wire is formed into a circle having a diameter of 11.1 cm and is placed in a uniform magnetic field of 2.79 mT. The wire carri

Ber [7]

Answer:

Maximum torque on the wire is $1.34\times 10^{-4}\ N-m$

Explanation:

It is given that,

Diameter of the wire, d = 11.1 cm = 0.111 m

Radius of wire, r = 0.0555 m

Magnetic field, $B=2.79\ mT=0.00279\ T$

Current, I = 5 A

We need to find the maximum torque on the wire. Torque is given by :

$\tau=IAB\ sin\theta$

Torque is maximum when, $\theta=90$

$\tau=IAB$

$\tau=5\times \pi \times (0.0555)^2\times 0.00279$

$\tau=0.000134\ N-m$

or

$\tau=1.34\times 10^{-4}\ N-m$

So, the maximum toque on the wire is $1.34\times 10^{-4}\ N-m$ . Hence, this is the required solution.

6 0

3 years ago

While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s

yawa3891 [41]

Answer:

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Explanation:

Let L represent Moe's height during the leap.

Moe's velocity v at any point in time during the leap is;

v = dL/dt = u - gt .......1

Where;

u = it's initial speed

g = acceleration due to gravity on Mars

t = time

The determine how far the cricket was off the ground when it became Moe’s lunch.

We need to integrate equation 1 with respect to t

L = ∫dL/dt = ∫( u - gt)

L = ut - 0.5gt^2 + L₀

Where;

L₀ = Moe's initial height = 0

u = 105m/s

t = 56 s

g = 3.75 m/s^2

Substituting the values, we have;

L = (105×56) -(0.5×3.75×56^2) + 0

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

5 0

3 years ago

How do the gravitational potential energy and the kinetic energy of the ball change as the ball rolls down the ramp?

belka [17]

Explanation:

The gravitational potential energy is given by :

P = mgh

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

As the ball reaches the bottom of the ramp, its potential energy decreases and kinetic energy increases.

It imply that, when the ball at the top most height, its gravitational potential energy is maximum and zero kinetic energy and when ball reaches the bottom of the ramp, it will have maximum kinetic energy and zero potential energy.

3 0

3 years ago

What is an astronomical unit (AU)?

avanturin [10]

The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.

Now it's defined as 149,597,870,700 meters exactly.

4 0

3 years ago

You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the

Jet001 [13]

Answer:

Explanation:

Given

initial velocity $u=11.2\ m/s$

At maximum height velocity of ball is zero

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)^2 -(11.2)^2=2\times (-9.8)\times (s)$

$s=\frac{11.2^2}{2\times 9.8}$

$s=6.4$

time taken by the ball to reach the maximum height

$v=u+at$

$0=11.2-9.8\times t$

$t=\frac{11.2}{9.8}$

$t=1.142\ s$

At $t=2\ s$ height of ball is

$h=ut+\\frac{1}{2}at^2$

$h=11.2\times 2-\frac{1}{2}9.8\times (2)^2$

$h=22.4-19.6$

$h=2.8\ m$

i.e. ball is moving downward

height at $v=5\ m/s$

$v^2-u^2=2as$

$s=\frac{25-125.4}{2\times (-9.8)}$

$s=5.12\ m$

8 0

3 years ago