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navik
1 month ago
10

Three cars are tested in a 16 m/s frontal crash. The results of the crash tests are shown below, with data indicating how much t

he crumple zone and safety cell of each car deformed. In each car the dummy was secured by the same seat belt and airbag. Which car did the best job of protecting the dummy?

Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0

Response:

  Vehicle C

Rationale:

The deformed crumple zone in Vehicle C is considerably lengthier, resulting in a reduced average deceleration. This lower rate of acceleration exerts less force on the dummy. Vehicle C provided superior protection for the dummy.

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A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
inna [3103]

Response:

The car's acceleration magnitude is 35.53 m/s²

Details:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car during the collision = a_c

Using Newton's third law of motion;

The force exerted by the truck equals the force exerted by the car.

The car's force acts in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Thus, the car's acceleration magnitude is 35.53 m/s²

3 0
2 months ago
The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Sav [3153]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
2 months ago
If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,
Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0
1 month ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
ValentinkaMS [3465]
The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height. 

<span>In this scenario, you can demonstrate it as follows: </span>

<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>

<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>

<span>the total flight duration is the period required to return to a height of 0 m, thus </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time =? </span>
<span>a = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now let's examine the horizontal distance. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range =? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>therefore </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 month ago
Starting with only the Balmer series light (visible light), how could we ensure that the solar panels generate a current that Ma
ValentinkaMS [3465]

The right answer is (a).

Solar panels create electric current through the photoelectric effect, which describes how photons strike certain material surfaces, resulting in the release of electrons when light with the correct frequency hits them. A photon will interact with an electron on the panel, causing it to be ejected from the panel's surface.

As the illumination on the panel becomes brighter, the intensity of the light rises, indicating an increase in the number of photons. Each photon has the potential to liberate an electron; thus, as the number of incoming photons rises, so does the quantity of freed electrons. Given that the photoelectric current reflects the rate at which these electrons flow, an increase in light intensity leads to a corresponding rise in the photoelectric current.

If the frequency of the light is increased without a change in brightness, the photoelectric current remains the same because the total number of photons does not increase. Yet, the electrons that are ejected do escape with higher kinetic energy. However, since the total number of electrons liberated stays unchanged, the current remains constant regardless of the electrons' increased energy. Thus, option b is incorrect.

Increasing the wavelength of the light means the energy of the photons decreases. This would cause the emitted electrons to have lower energy. However, if the brightness is consistent, the number of electrons remains the same, and as a result, there would be no change in the photoelectric current. Therefore, choice (c) is also incorrect.

The correct answer is (a). To generate the needed current, the brightness of the incident light must be increased.

8 0
2 months ago
Read 2 more answers
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