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Lerok [7]
2 years ago
5

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds

. By how much would the force have been reduced if the car had had a crumple zone that increased the collision time to 2.2 seconds? Give your answer to the nearest whole number.
Physics
1 answer:
horrorfan [7]2 years ago
4 0

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

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A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
2 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
2 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta
Dvinal [7]
Nobody will do that for 5 points loll
8 0
2 years ago
1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?
Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

<h3>4.53 kgm/s</h3>

Hope this helps you

4 0
2 years ago
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