Can force be two-dimensional?

The application of solid-state physics that is the study of the arrangement of atoms in a solid is

balandron [24]

Answer:

crystallography.......

8 0

3 years ago

A 14000N car traveling at 25m/s rounds a curve of radius 200m. Find the following: a. The centripetal acceleration of the car.

tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car $W=14,000\ N$

mass of car $m=\frac{14,000}{9.8}=1428.57\ N$

velocity of car $v=25\ m/s$

radius $r=200\ m$

(a)Centripetal acceleration is given by

$a_c=\frac{v^2}{r}$

$a_c=\frac{25^2}{200}$

$a_c=3.125\ m$

(b)Force that provide centripetal acceleration

$F=F_c=\frac{mv^2}{r}$

$F=\frac{1428.57\times 25^2}{200}$

$F=4464.285\ N$

(c)Friction force between car and tires is given by

$=\mu N$

where $\mu$ =coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

$F_c=F_r$

$4464.285=\mu \times 1428.57\times 9.8$

$\mu =0.318$

6 0

3 years ago

Read 2 more answers

What are the different intensity of an earthquake

Masja [62]

Ranking Earthquake Intensity

Magnitude Average number per year Modified Mercalli Intensity

2.0 – 2.9 >1 million I

3.0 – 3.9 about 100,000 II – III

4.0 – 4.9 about 10,000 IV – V

5.0 – 5.9 about 1,000 VI – VII

Explanation:

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5 0

3 years ago

5. A race car has a mass of 710 kg. It starts from rest and travels 40.0m in 3.0s. The car is uniformly accelerated during the e

Lana71 [14]

5. 6319 N

First of all, we need to find the acceleration of the car, which can be found by using the equation

$S=\frac{1}{2}at^2$

where

S = 40.0 m is the distance travelled by the car

t = 3.0 s is the time taken

a is the acceleration

Solving for a, we find

$a=\frac{2S}{t^2}=\frac{2(40.0 m)}{(3.0 s)^2}=8.9 m/s^2$

So now since we know the mass of the car, m=710 kg, we can find the net force acting on the car:

$F=ma=(710 kg)(8.9 m/s^2)=6319 N$

6. 62.5 m

In this case, we know the breaking force applied on the car,

$F=-5000 N$

(with a negative sign since its direction is opposite to the car's motion)

and the mass of the car

$m=1000 kg$

so we can find its acceleration:

$a=\frac{F}{m}=\frac{-5000 N}{1000 kg}=-5 m/s^2$

So now we can find the minimum distance to stop by using the equation

$v^2-u^2 = 2ad$

where in this case we have

v = 0 m/s is the final speed

u = 25 m/s is the initial speed

a = -5 m/s^2 is the acceleration

d is the distance

solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(25 m/s)^2}{2(-5 m/s^2)}=62.5 m$

7a. 14 m/s

We can solve the problem by using the law of conservation of energy: in fact, the initial gravitational potential energy of the person is all converted into kinetic energy as she hits the water below

$mgh=\frac{1}{2}mv^2$

where

m = 65 kg is the mass of the person

g = 9.8 m/s^2

h = 10 m is the initial height of the diver

v is the final speed as she enters the water

Solving for v, we find

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s$

7b. -3185 N

We need to find the acceleration of the diver during the motion of 2.0 m below the water:

$v^2-u^2 = 2ad$

where

v = 0 is the final speed

u = 14 m/s is the initial speed as she enters the water

a is the acceleration

d = 2.0 m is the distance covered

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(14 m/s)^2}{2(2.0 m)}=-49 m/s^2$

And so now we can find the net force acting on the diver

$F=ma=(65 kg)(-49 m/s^2)=-3185 N$

8a. -133 m/s^2

The acceleration of the passenger is given by

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

where

v = 0 m/s is the final speed

u = 13.3 m/s is the initial speed

t = 0.10 s is the time interval

Solving for a, we find

$a=\frac{0-13.3 m/s}{0.10 s}=-133 m/s^2$

8b. 3325 N

The force that the driver must exert on the child is equal in magnitude to the force experienced by the child during the stop of the car, so

$F=ma$

where

m = 25 kg is the mass of the child

$a=-133 m/s^2$ is the acceleration

So, the magnitude of the force will be

$F=(25 kg)(133 m/s^2)=3325 N$

8c. 245 N

The weight of the child is given by

$W=mg$

where

m = 25 kg is the child's mass

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation,

$W=(25 kg)(9.8 m/s^2)=245 N$

8d. 55.1 lb

Since we know that

$1 lb = 4.45 N$

We can find the weight in pounds by setting the following proportion

$1 lb : 4.45 N = x : 245 N$

Solving for x,

$x=\frac{(1 lb)(245 N)}{4.45 N}=55.1 lb$

8e. Chances are very low

The force that the driver should exert on the child is

F = 3325 N

This force is equivalent to the force required to lift an object of mass m:

$F=mg\\m=\frac{F}{g}=\frac{3325 N}{9.8 m/s^2}=339.3 kg$

So, it is equivalent to the force required to lift an object of 339.3 kg, which is quite a lot. therefore, the changes are very low.

5 0

3 years ago

Energy is dissipated from a rollercoaster cart as it mo es along the track.

aliya0001 [1]

What’s is as it mo es along track what does that mean more explanation

6 0

3 years ago