Which years had the least sunspot activity?

The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The aver

Crank

Answer:

The potential difference is 121.069 V

Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, $\rho = 5.5\ \Omega-m$

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

$\rho = R\frac{A}{l}$

$R = \rho \frac{l}{A}$

$R = 5.5\times \frac{1.4}{6.36\times 10^{- 3}} = 1210.69\ \Omega$

Now, using Ohm's Law:

V = IR

$V = 0.1\times 1210.69 = 121.069\ V$

4 0

3 years ago

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0

3 years ago

11. एक समान चुम्बकीय क्षेत्र में लम्बवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की

mojhsa [17]

Answer:

Explanation:

If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule

to derive the radius of the path of the charge we apply

F= mv^2/r=Bev

where

m= mass of the electronic charge

e=charge

B=magnetic field

v=average speed

r=radius

rearranging we have

r=mv^2/Bev

r=mv/Be

5 0

3 years ago

A Imagine you derive the following expression by analyzing the physics of a particular system: v2 + 2az. The problem requires so

Nuetrik [128]

Answer:

z = 93.2 m

Explanation:

We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration

v² = v₀² + 2 a d

If we make a term-to-term comparison with the expression obtained, they are equivalent

u² = v² + 2 a z

From here we can clear the position

2 a z = u² –v²

z = (u² –v²) / 2 a

Let's calculate

For the speed to reduce the acceleration must be negative

z = (0 - 21.8²) / 2(- 2.55)

z = 93.2 m

7 0

2 years ago

An engine pulls a train of 20 freight cars, each having a mass of 4.900 × 104 kg, with a constant force. The cars move from rest

bixtya [17]

Answer:

The force with which the tenth car pulls the eleventh one is called tension and is equal to:

T=119715.91 N

Explanation:

The force (F) with which the tenth car pulls the eleventh one is called tension and its direction is the X-direction or horizontal. According to Newton's Second Law of motion:

$\sum F=ma$

That is, the force of the car is equal to the acceleration (a) times its mass (m). The acceleration is the change in the velocity divided by the time (i is for initial and f is for final).

$a=\frac{v_f-v_i}{t_f-t_f}$

Using Newton's second law:

To find the forces, you have to solve the equilibrium in X-direction:

$\sum F_x=T=ma_x$

Now you can substitute the accelertion in terms of velocity and time:

$\sumF_x=T=ma_x=m\frac{v_f-v_i}{t_f-t_i}$

Solve the equation using the data from the problem, remember that the mass of the object is 10 times the mass of one car because the 10th car has to pull all the other cars:

$T=m\frac{v_f-v_i}{t_f-t_i}=(10)*(4.900 \times 10^4 kg)(\frac{4.3 m/s}{17.60s})\\T=119715.91 N$

4 0

3 years ago