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2 years ago

Inertia keeps your vehicle moving until a force slows it down, like __________.

2 answers:

6 0

Certainly 'A' and 'C'.

And I don't see why 'B' couldn't also exert a force against the car
that would slow it down.

I'd say "all of the above".

GrogVix [38]2 years ago

5 0

Technically all 3 of those would be correct answers. But applying the brakes would probably be the correct answer if all of the above isn't a answer.

You might be interested in

The potential in a region of space due to a charge distribution is given by the expression V = ax2z + bxy − cz2 where a = −3.00

Elis [28]

Answer:

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

Explanation:

We know that the relationship between the electric field $\vec{E}(\vec{r})$ and the potential $V(\vec{r})$ is given by

$\vec{E} ( \vec{r}) = - \vec{\nabla} V(\vec{r})$

So, for our potential:

$V(r) = a x^2 z + b x y - c z^2$

the electric field is :

$\vec{E} ( \vec{r}) = - \vec{\nabla} ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x}, \frac{ \partial }{\partial y} , \frac{ \partial }{\partial z}) ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial y} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial z} ( a x^2 z + b x y - c z^2 ))$

$\vec{E} ( \vec{r}) = - ( 2 a x z + b y , b x , a x^2 - 2 c z )$

This is the our electric field. At vector point

$\vec{r} = (0, -8.00 \ m, - 8.00 \ m)$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 2 a * 0 * (-8.00 \ m) + b (-8.00 \ m) , b * 0 , a 0^2 - 2 c (-8.00 \ m) )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 0 - b 8.00 \ m , 0 , 0 + 2 c 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( b 8.00 \ m , 0 , - 2 c 8.00 \ m )$

Knowing

$b= 6.00 \frac{V}{m^2}$

and

$c=9.00 \frac{V}{m^2}$

the electric field is

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 6.00 \frac{V}{m^2} * 8.00 \ m , 0 , - 9.00 \frac{V}{m^2} * 2* 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

7 0

2 years ago

You are taking a turn at 30.0 m/s on a ramp of radius 25.0 m. What is your acceleration?

Nat2105 [25]

Acceleration = v^2 / r

Acceleration = 30^2/25

Acceleration = 900/25

Acceleration = 36 m/s^2

6 0

2 years ago

Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

34kurt

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

$\mathsf{v^2 = u^2 + 2as}$

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

$\mathsf{v^2 - u^2 =2as }$

$\mathsf{a = \dfrac{v^2 - u^2 }{2s}}$

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

$\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}$

$\mathsf{a = \dfrac{- 771.7284 }{2}}$

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

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

$\mathsf{t = \dfrac{-27.78}{-385.86}}$

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

3 0

3 years ago

The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo

spin [16.1K]

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

$W=\vec F\ \vec r$

It can be written in its scalar version as

$W=F.d.cos\theta$

Being F and d the magnitudes of the force and displacement, and $\theta$ the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

$W=N.d.cos90^o=0$

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

$W=F_r(0)cos180^o=0$

3 0

3 years ago

A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri

frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0

3 years ago

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