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2 years ago

12

__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.

1 answer:

Rainbow [258]2 years ago

5 0

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

djyliett [7]

I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
$W_f=F_f\cdot L$
So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
$B: mgh_2+\frac{mv_2}{2}$
As said before this energy must be the same as the energy of a car approaching the loop:
$mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL$
Now we solve the equation for $v_1$ :
$v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}$

4 0

2 years ago

Read 2 more answers

if a net horizontal force of 175 N is applied to a bike whos mass is 43 kg what acceleration is produced

Anna [14]

Explanation:

f=175N

m=43kg

a=?

know

f=ma

a=f/m

a=175/43

a=4.06m/s

3 0

2 years ago

89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a

ddd [48]

Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

$v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05$

v = 871779.788 m/s

or

$v=8.71\times 10^5\ m/s$

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

$t=\dfrac{v-u}{a}$

$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

Hence, this is the required solution.

4 0

2 years ago

A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro

LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock = 0.22kg

Initial velocity = 20m/s

Distance moved = 10m

Unknown:

Initial kinetic energy of the rock = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2}$ x 0.22 x 20² = 44J

6 0

1 year ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago