Ask question

Ask question

All categories

Karo-lina-s [1.5K]

1 year ago

10

A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto

ps the lorry in 1.5 seconds. What was the change in momentum?

1 answer:

zysi [14]1 year ago

7 0

Answer:

3000 kg.m/s

Explanation:

Momentum, p is a product of mass and velocity hence

p=mv where m is mass and v is velocity.

Change in momentum is given by $m(v_f-v_i)$ where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then

Change in momentum= 6000(0-0.5)=-3000 kg.m/s

You might be interested in

A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increa

Harlamova29_29 [7]

Answer:

T₂ = 111.57 °C

Explanation:

Given that

Initial pressure P₁ = 9.8 atm

T₁ = 32°C = 273 + 32 =305 K

The final pressure P₂ = 11.2 atm

Lets take the final temperature = T₂

We know that ,the ideal gas equation

If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$T_2=\dfrac{P_2}{P_1}\times T_1$

Now by putting the values in the above equation ,we get

$T_2=\dfrac{11.2}{9.8}\times 305\ K$

$T_2=348.57\ K$

T₂ = 384.57 - 273 °C

T₂ = 111.57 °C

3 0

2 years ago

A transverse wave is traveling from north to south. Which statement could be true for the motion of the wave particles in the me

Vikki [24]

Transverse waves travel on a direction that is perpendicular to the motion of the particles (or whatever medium is waving) So the particles must be moving east to west, which is transverse to the north-south motion of the wave

3 0

1 year ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

A spring stretches 0.220 m when a 0.400 kg-mass is hung from it. What is its spring constant? (Mass is not a force )

Fantom [35]

We want to know the amount of force that stretches the spring 0.22 m.
That force is the WEIGHT of the mass hung from it.
The weight of the mass is (mass) times (gravity).
To do that calculation, we need to know the value of gravity, but
gravity has different values on every planet. I shall assume that
this whole springy question is taking place on Earth, so that the
value of gravity is 9.8 m/s² .

The weight of the mass is (0.4 kg) x (9.8 m/s²) = 3.92 Newtons.

The spring constant is

(force/length of the stretch)

= (3.92 Newtons) / (0.22 meters)

= (3.92 / 0.22) Newtons/meter

= 17.82 N/m .

8 0

1 year ago

Read 2 more answers

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

andrey2020 [161]

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field $U=4\times 10^{-6}J$

Value of inductance L =0.060 H

Energy stored in magnetic field is given by $U=\frac{1}{2}Li^2$

$i=\sqrt{\frac{2U}{L}}$

$i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A$

So the current flowing through rectangular toroid will be 0.01154 A

3 0

1 year ago