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mars1129 [50]
2 years ago
6

A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o

f the initial acceleration vector and compare the initial acceleration vector’s magnitude with respect to the acceleration magnitude at the moment of the crash. [*** DRAW A DIAGRAM ***]
Physics
1 answer:
earnstyle [38]2 years ago
5 0

Answer:

the direction of acceleration of the vehicle is the same direction of its velocity of car

s acceleration has the opposite direction to the car speed.

Explanation:

The initial acceleration of the car can be calculated with

          v = v₀ + a t

          a = (v-v₀) t

       

indicate that the initial velocity is zero (v₀ = 0 m / s)

         a = v / t

         a = 300 / t

the direction of acceleration of the vehicle is the same direction of its acceleration movement.

When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small

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An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc
Afina-wow [57]
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity. 

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

Distance = 10 * 1.06
Distance = 10.6 meters

The boat should be 10.6 meters away horizontally from the point of release.
4 0
2 years ago
Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?
morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays,  parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

3 0
2 years ago
Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli
Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: sin \theta=\frac{m\lambda}{\alpha}

Given m=3,\lambda=6.33\times 10^-^7 and \alpha=0.00015:

#Substitute our variable values in the minima equation to obtain \theta:

\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad

#draw a triangle to find the relationship between \theta, y \ and L.

tan(\theta)=y/L               #where y=4.05cm

L=y/tan(\theta)=3.20

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

sin \theta=\frac{m\lambda}{\alpha}, where m=4 given that it's the minima with the smallest distance.

sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad

#we then use tan(\theta)=y/L to calculate L=4.5cm

Then from the equation subtract y_3 from y:

4.50cm-4.05cm=0.45cm

Hence, the distance \bigtriangleup y is 0.45cm

8 0
2 years ago
A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o
faltersainse [42]

m = mass = 5 kg

v_{i} = initial velocity = 100 m/s

v_{f} = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m(v_{f} - v_{i})

30 = 5 (v_{f} - 100)

v_{f} = 106 m/s

5 0
2 years ago
Read 2 more answers
A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
Tems11 [23]

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

= 100 x 1

= 100 pound ft.

For equilibrium

Torque of F = torque of weight.

F x 2  = 100

F = 50 pounds .

7 0
2 years ago
Read 2 more answers
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