Ask question

Ask question

All categories

15 days ago

10

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

e rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.

You might be interested in

Two carts are involved in an elastic collision. Cart A with mass 0.550 kg is moving towards Cart B with mass 0.550 kg, which is

Sav [3153]

Answer:

A) $v_b=0.8\ m.s^{-1}$ denotes the resultant velocity of cart B post-collision.

B) $KE_A=0.176\ J$

C) $KE_B=0\ J$

D) $ke_a=0\ J$

E) $ke_B=0.176\ J$

F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.

G) Yes, momentum is conserved in every elastic collision.

Explanation:

Given:

mass of car A, $m_a=0.55\ kg$

mass of car B, $m_b=0.55\ kg$

initial velocity of car A, $u_a=0.8\ m.s^{-1}$

final velocity of car A, $v_a=0\ m.s^{-1}$

A)

The question mentions the cars experience an elastic collision:

By applying momentum conservation principles:

$m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b$

$0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b$

$v_b=0.8\ m.s^{-1}$ denotes the resulting velocity of cart B after collision.

B)

Initial kinetic energy of cart A:

$KE_A=\frac{1}{2} m_a.u_a^2$

$KE_A=0.5\times 0.55\times 0.8^2$

$KE_A=0.176\ J$

C)

Initial kinetic energy of cart A:

$KE_B=\frac{1}{2} \times m_b.u_b^2$

$KE_B=0.5\times 0.55\times 0^2$

$KE_B=0\ J$

D)

The final kinetic energy of cart A:

$ke_A=\frac{1}{2} m_a.v_a^2$

$ke_a=0.5\times 0.55\times 0^2$

$ke_a=0\ J$

E)

The final kinetic energy of cart B:

$ke_B=\frac{1}{2} m_b.v_b^2$

$ke_B=0.5\times 0.55\times 0.8^2$

$ke_B=0.176\ J$

F)

Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.

G)

Indeed, momentum is consistently conserved in elastic collisions.

5 0

1 month ago

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Yuliya22 [3333]

The time required for the water balloon to reach the ground is given as

$h = \frac{1}{2} gt^2$

Here we understand that

$h = 18 m$

$g = 9.8 m/s^2$

Now applying the earlier mentioned formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

Now in the same time frame, we can conclude the distance covered will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

Thus, it will land at a distance of 15.7 m from where it started

5 0

2 months ago

Un tren parte de la ciudad A, a las 8 h. con una velocidad de 50 km/h, para llegar a la ciudad B a las 10 h. Allí permanece dura

Keith_Richards [3271]

Response:

AB = 100 km; BC = 80 km; AC = 180 km

Time of arrival = 11:30

Reasoning:

1. Distance from A to B

(a) Duration of travel

Duration = 10:00 - 8:00 = 2.00 hours

(b) Distance

Distance = speed × time = 50 km/h × 2.00 h = 100 km

2. Distance from B to C

Distance = 80 km/h × 1 h = 80 km

3. Summary of Distances

AB = 100 km

BC = 80 km

AC = 180 km

4. Time of Arrival

Departure from A = 08:00

Travel duration to B = 2:00

Arrival at B = 10:00

Waiting time at B = 0:30

Departure from B = 10:30

Travel duration to C = 1:00

Arrival at C = 11:30

8 0

3 months ago

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

3 months ago