All categories

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

Physics

1 answer:

yan [13]3 years ago

5 0

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

You might be interested in

I'm not an idiot so I'm pretty sure c and d are not the answers. I'm confused because don't they both do this. As it picks up sp

krok68 [10]

The answer is A because it's MOVING

ithink

8 0

2 years ago

Read 2 more answers

What happens to the electric potential energy of a negatively charged ion as it moves through the water from the negative probe

Nataly [62]

Answer:

Decreases.

Explanation:

Electric potential energy is the potential energy which is associated with the configuration of points charge in a system and it is the result of conservative coulomb force.

When the negatively charge ion is at the position of the negative probe than its potential energy is positive when it is move towards the positive probe it's potential energy becomes negative due to the negative ion.

Therefore, potential energy is decreases when negative charge ion moves through the water from negative probe to positive probe.

5 0

3 years ago

A bungee jumper feels weightless as she falls toward the earth.what accounts for her weightless feeling when she fall freely?1.

Ann [662]

The feel of weight comes due to the normal reaction force given by the support. Hence, the condition of weightlessness is when the normal reaction force becomes zero. So, during free fall there is no support which can provide the normal reaction. Hence, the bungee jumper feels weightless as she falls towards the earth because of the lack of support force that balances gravity.

Hence, the answer is 3.

8 0

3 years ago

Read 2 more answers

In what direction must the current I2 flow in the straight wire in order to create a field at the center of the loop in the dire

guajiro [1.7K]

Answer:

Right to left

Explanation:

From right hand thumb rule, the magnetic field at the centre of loop points out of the plane of paper. Thus the straight wire current should create magnetic field into the plane of paper at the centre. Hence the direction of current in long wire should be from right to left

6 0

3 years ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by

∑F = ma (1)

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:

a_max = -w^2A (2)

3- The angular frequency w of a wave is related to the frequency f by:

w = 2π f (3)

Given Data

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>

Solution

(a)

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:

∑F_y -mg= ma

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So

0 = m(g + a)

a = —g

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):

a = —w^2A

where a = —g at the highest point. So

g = w^2A

Solve for w:

w =√g/A

Substitute for w from Equation (3):

2πf = √g/A

Solve for f :

f = 1/2π√g/A

Substitute numerical values:

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0

3 years ago