A car with a mass of 2.0 * 10^3 kg is traveling at 15 m/s. what is the momentum of the car?

How early do you think i should start saving and searching for scholarships and state aid

brilliants [131]

Never too early to start searching. Do some research about student savings versus parent savings though. If a student has savings, they will make you use it to pay for college, while the same amount of savings in the parents name may be exempt. Check it out.

8 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached

RUDIKE [14]

Answer:

Wmax = 63.65 ≈ 64 lb

Explanation:

6 0

3 years ago

A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

Alex73 [517]

Answer:

$V_f = 13.8863 \angle 60.89\°$

Explanation:

Our values are,

$m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s$

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

$m_1v_1=(m_1+m_2)v_fcos\theta$

For the particular case on the Y axis, we do the same with the speed of object 1.

$m_2v_2=(m_1+m_2)v_fsin\theta$

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

$Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°$

Replacing in any of the two functions, given above, we will find the final speed after the collision,

$(950)(16)=(950+1300)V_fcos(60.89)$

$V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}$

$V_f = 13.8863 \angle 60.89\°$

8 0

3 years ago

Answer the following questions.

Sever21 [200]

Answer:

1. B 2. A. 3. A. 4. B 5.B

Explanation:

1) The statement is false, because the acceleration (by definition) is the rate of change of velocity, which is a vector.

As a vector, it can change the magnitude, the direction, or both. As the speed is only the magnitudde, there can be an acceleration at constant speed, as in an uniform circular movement, where the acceleration produces a change in direction, not in the speed.

2) If the body is in motion (assuming that it is only translational motion) the position (displacement from the origin) must be change with time, so the statement is true.

3) This is true, as the acceleration, as we said above, is the rate of change of the velocity vector.

4) The statement is false, as an object moving at constant speed along a straight line is in movement, and is not accelerated at all.

5) The statement is false, as a body can momentarily come to an stop (before changing direction, for instance), and be accelerated at the same time.

The best example is an object thrown up in the air: once released, the only force acting on it is gravity, which produces an acceleration in opposite direction to the initial speed, until it causes the object to stop, change direction, and finally fall freely.

3 0

3 years ago