(I will give brainliest whoever helps me !!)

A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago

Technician A says that a power-assisted brake system reduces stopping distances compared to a nonpower-assisted brake system. Te

yanalaym [24]

Answer:

Technician B

Explanation:

here on analyzing both the statements from technician A and technician B. The Statement from Technician B is more logical and correct. That the power-assisted brake system reduces the force that the driver must exert on the brake pedal.

The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.

4 0

2 years ago

A person is walking briskly in a straight line. The figure shows a graph of the person’s position x as a function of time t. Wha

Nina [5.8K]

The average velocity over the interval $6\le t\le 10$ is

$\dfrac{x(10)-x(6)}{10\,\mathrm s-6\,\mathrm s}=\dfrac{6\,\mathrm m-2.67\,\mathrm m}{4\,\mathrm s}=0.8325\,\dfrac{\mathrm m}{\mathrm s}$

where $x(t)$ is the person's position at time $t$ . I've taken the liberty of estimating $x(6)\approx2.67\,\mathrm m$ . The closest choice among the answers is $0.8\,\dfrac{\mathrm m}{\mathrm s}$ .

7 0

2 years ago

A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the secon

Gemiola [76]

The average speed of a moving object is the rate of change of a certain distance with respect with time. It is equal to the total distance that was traveled by the object over the total time it takes to travel that distance. For this problem we need to assume that the total distance that was traveled would be equal to 120 miles. So, for the first half of the distance or 60 miles at a speed of 30 miles per hour, the time taken would be two hours. For the remaining 60 miles at a speed of 60 miles per hour, 1 hour is total time traveled. So, we calculate the average speed as follows:

Average speed = total distance / total time
Average speed = 120 miles / 2 hr + 1 hr
Average speed = 40 mi / hr

5 0

3 years ago

Question 2 Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid

Mama L [17]

Answer:

The answer is 12.27 g (NH4)3PO4

Explanation:

Step 1: balance the chemical equation:

H3PO4 + 3NH3 → (NH4)3PO4

step 2: the molar masses of each of the reagents and products are calculated using the periodic table:

H3PO4:

3 atoms of H: 3x1=3 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=3+30.97+64=97.97 g/mol

3NH3:

3 atoms of N: 3x14=42 g/mol

9 atoms of H: 9x1=9 g/mol

Molar mass=42+9=51 g/mol

(NH4)3PO4:

3 atoms of N: 3x14=42 g/mol

12 atoms of H: 12x1=12 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=42+12+30.97+64=148.97 g/mol

we make a rule of three to calculate the amount of ammonium phosphate:

51 g NH3------------------148.97 g (NH4)3PO4

4.2 g NH3-----------------x g (NH4)3PO4

Clearing the x, we have:

$x g (NH4)3PO4 = \frac{(4.2 g NH3)x(148.97 g (NH4)3PO4)}{51 g NH3}=12.27 g (NH4)3PO4$

7 0

3 years ago