How are the magnetic field lines near the south pole of a magnet affected when a second south pole is brought near it?

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi

Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

$E_A} =E_{B}$

Spring potential energy =potential energy due to elevation

0.5*k*x²= mg $(h_{B}-h_{A} )$ =mgh

0.5*k*2.3²= 430*9.81*6

k=9568.92 N/m

For safety reason

k"=1.13 *k= 1.13*9568.92

k"=10812.88 N/m

agsin from conservation of energy

$E_A} =E_{C}$

spring potential energy=change in kinetic energy

0.5*k"*x²=0.5*m* $V_{max}^{2}$

10812.88 *2.3²=430* $V_{max}^{2}$

$V_{max}$ =11.53 m/s

5 0

3 years ago

A roller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. From there it starts into a

Softa [21]

Answer:

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

$F_N = F_C - F_G = m\frac{v^{2} }{r} - mg$

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

$E_{tot}=mgh + \frac{1}{2} mv^{2}$

At the top of the hill:

$E_{tot}= mgh_{hill}$

At the top of the loop:

$E_{tot}=mgh_{loo}_p +\frac{1}{2} m v^{2}$

Setting both energies equal and canceling the mass m gives:

$gh_{hill} = gh_{loo}_p + \frac{1}{2} v^{2}$

Solving for v:

$v^{2} = 2g(h_{hill}-h_{loo}_p)$

Using v in the first equation:

$F_N = \frac{2mg(h_{hill}-h_{loo}_p)}{r} - mg$

$F_N = 955.5N$

7 0

3 years ago

Two objects separated by a distance r are each carrying a charge q The magnitude of the force exerted on the second object by th

Sindrei [870]

Answer:F=4F

Explanation: Columbs law states that The force between the two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them

Force between the two charges is given by

F=K*q1*q2/r^2

if one charge become 4 times, new force is,

F=4(K*q1*q2)/r^2

F=4F

Where q1 and q2 are the point charges

r is the distance between the two charges

K is a constant of proportion called electrostatic force

4 0

3 years ago

14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball

Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

Height h = 75 m

Initial velocity u = 0 ( vertical velocity )

Acceleration due to gravity g = 9.8 m/s²

Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

#SPJ1

7 0

1 year ago

When you and a friend move a couch to another room you exert a force of 75 N over 5m how much work will you do

sesenic [268]

The work done is 375 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this problem,

F = 75 N is the force applied to the couch

d = 5 m is the displacement

Assuming the force applied to the couch is parallel to the motion, $\theta=0$

And so, the work done is

$W=(75)(5)(cos 0)=375 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

6 0

3 years ago