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Dahasolnce [82]
3 years ago
8

Using your knowledge of momentum explain why cars are designed to have crumple zones?

Physics
1 answer:
Molodets [167]3 years ago
4 0

Answer: I think cars are designed to have crumble zone because lets say you're going 60-70 mph and you hit a brick wall that cant move, it would be a very hard jolt causing the beings inside to get thrown forward, but if it has a crumble zone it would slow the the jolt from is slowing down in the hit.

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a car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s? for 15 s. At the end of this time,
Firlakuza [10]

V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)

V = 15 + 2 X 15

V = 45 m/s

6 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig
Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = \frac{1}{3}\pi r^2 h

thus,

volume of the cone = \frac{1}{3}\pi 1.2^2\times 4 = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the \frac{1}{4}times the total height

thus,

center of mass lies at,  h' = \frac{1}{4}\times4=1m

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0
3 years ago
A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.
nikdorinn [45]

Answer:

a

 \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

b

 I = 0.106 \  A

Explanation:

From the question we are told that

  The current is  I =  0.106 \  A

   The length of one side of the square a = 4.60 \  cm = 0.046 \  m

    The separation between the plate is  d = 4.0 mm  = 0.004 \ m

Generally electric flux is mathematically represented as

       \phi_E = \frac{Q}{\epsilon_o}

differentiating both sides with respect to t is  

       \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}

=>     \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} *I

Here \epsilon_o is the permitivity of free space with value  

        \epsilon _o  =  8.85*10^{-12} C/(V \cdot m)

=>   \frac{d \phi_{E}}{dt}  = \frac{0.106}{8.85*10^{-12}}

=>   \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

Generally the displacement current between the plates in A

    I = 8.85*10^{-12} * 1.1977 *10^{10}

=>  I = 0.106 \  A

 

3 0
2 years ago
fired, the ball moves 15.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.
sergeinik [125]

Answer:

Explanation:

Where is the remaining part of the question? This is a very easy one

6 0
2 years ago
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