All categories

3 years ago

Which accident will be more damaging, collision between two trucks

2 answers:

4 0

It is collision truck because the more is the mass, more is the inertia and therefore more is the momentum. Mass of the trucks is more than that of cars so collision of trucks will cause more damage.

Vitek1552 [10]3 years ago

3 0

Answer:

collision between truck

Explanation:

Answer. Collision between trucks, because more is the mass, more is the inertia and therefore more is the momentum. Mass of the trucks is more than that of cars so collision of trucks will cause more damage.

You might be interested in

A 51.0 g stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is 16.0 cm/s

Nina [5.8K]

Answer:

A) 32.22 N/m b) 0.0156 m c) 4 Hz

Explanation:

Using Hooke's law;

T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.

M = 51 g = 51 / 1000 in kg = 0.051kg

Make k subject of the formula

T/2π = √m / k

Square both sides

T^2 / 4π^2 = m/k

Cross multiply

K = 4 π^2 * m/T^2

K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m

B) using Hooke's law;

F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude

F= weight of the body = mass * acceleration due to gravity = 0.051*9.81

0.5 = 32.22 * e

e = 0.5/32.22 = 0.0156 m

C) frequency is the number of cycle completed in a second = 1 / period

F = 1 / 0.25 = 4Hz

6 0

3 years ago

What is a wavelength and give me a sentence with that word

RoseWind [281]

<span> A wavelenght is the distance between successive crests of a wave, especially points in a sound wave or electromagnetic wave.
</span>The diode laser operates at a wavelength<span> of 410 nm.</span>

4 0

3 years ago

At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 43.4 m

Sauron [17]

Answer:30.08 ms

Explanation:

Given

time Constant $\tau =43.4 ms =\frac{L}{R}$

Also rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in inductor's magnetic Field

Energy stored in Inductor is $U_L=\frac{1}{2}Li^2$

rate of Energy storing $\frac{dU_L}{dt}=\frac{1}{2}L\cdot 2i\times \frac{di}{dt}----1$

Rate of Energy dissipation from resistor i.e. Power is given by

$\frac{dU_R}{dt}=i^2R-----2$

Equating 1 and 2

$Li\cdot \frac{di}{dt}=i^2R$

$L(\frac{di}{dt})=R(i)-----3$

i is given by $i=\frac{V}{R}(1-e^{-\frac{t}{\tau }})$

$\frac{di}{dt}=\frac{V}{L}e^{-\frac{t}{\tau }}$

substitute the value of $\frac{di}{dt}$ in 3

$L(\frac{V}{L}e^{-\frac{t}{\tau }})=R\cdot \frac{V}{R}(1-e^{-\frac{t}{\tau }})$

$e^{-\frac{t}{\tau }}=1-e^{-\frac{t}{\tau }}$

$2e^{-\frac{t}{\tau }}=1$

$e^{-\frac{t}{\tau }}=0.5$

$e^{-\frac{t}{43.4\times 10^{-3}}}=0.5$

$\frac{t}{43.4\times 10^{-3}}=0.693$

$t=30.08 ms$

4 0

3 years ago

If a male runner in the fourth starting position ran the 400 m race in 44.40 s, how would you

Mumz [18]

V=d/t=>v=400m/44.40s=>v=9m/s

3 0

3 years ago

A cannonball is launched horizontally off a 20 m high castle wall with a speed of 85 m/s.

Aneli [31]

Answer:

The ball will be in flight for 2.0 s before it strikes the ground.

The range of the cannonball is 170 m.

Explanation:

Hi there!

The position vector of the cannonball at a time t is given by the following equation:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the cannonball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference on the ground, at the edge of the wall so that the initial position vector is r0 = (0, 20) m.

Using the equation of the vertical component of the position vector r, we can find the time it takes the ball to reach the ground:

y = y0 + 1/2 · g · t²

When the cannonball reaches the ground, y = 0:

0 = 20 m - 1/2 · 9.8 m/s² · t²

-20 m / -4.9 m/s² = t²

t = 2.0 s

The ball will be in flight for 2.0 s before it strikes the ground.

Now, we can calculate the horizontal component of the position vector when the ball reaches the ground at t = 2.0 s (i.e. the range of the cannonball).

x = x0 + v0 · t (x0 = 0 because we placed the origin of our frame of reference at the wall).

x = v0 · t

x = 85 m/s · 2.0 s

x = 170 m

The range of the cannonball is 170 m.

3 0

3 years ago