A kilometer is a measure of an object’s

A car of mass 1000kg is travelling in an eastward ( x) direction with a constant speed of 20m/s. It then hits a stationary truck

Aleonysh [2.5K]

Answer:

4.33 m/s, 5 m/s

Explanation:

mass of car, m = 1000 kg

initial velocity of car, u = 20 m/s along east

mass of truck, M = 2 x mass of car = 2000 kg

initial velocity of truck, U = 0 m/s

After collision, the velocity of car is v and the direction is 30° North of east.

The velocity of truck is V and the direction is 60° South of east.

Use conservation of momentum along X axis.

m x u = m x v x Cos 30 + M x V x Cos 60

1000 x 20 = 1000 x v x 0.866 + 2000 x V x 0.5

20 = 0.866 v + V .... (1)

Use conservation of momentum along y axis

0 = m x v x Sin 30 - M x V x Sin 60

0 = 1000 x v x 0.5 - 2000 x V x 0.866

v = 3.464 V ..... (2)

Substitute the value of v in equation (1)

20 = 0.866 x 3.464 V + V

V = 5 m/s

v = 0.866 x 5 = 4.33 m/s

Thus, the final speed of car is 4.33 m/s and the final speed of truck is 5 m/s after the collision.

8 0

3 years ago

Please indicate how long each bar is in centimeters and millimeters.

tatuchka [14]

Answer:

1). 9 cm

90 mm

2). 94 cm

940 mm

3). 27 cm

270 mm

4). 87 cm

870 mm

8 0

2 years ago

Hi free ponits hehehehhehbhrgivudksjbtyuvwijfe

Vilka [71]

Answer:

woohoo! thanks

Explanation:

3 0

2 years ago

Read 2 more answers

A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of th

Tamiku [17]

Answer:

the angle is about 67.79 degrees

Explanation:

We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)

We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

$half\,\,max-height = \frac{v_{yi}^2}{4\,g}$

we can use this information to find the y component of the velocity at that height via the formula:

$v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}$

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

$1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\$

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

$tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o$

3 0

3 years ago

A pesky rabbit has been feeding on Mrs. Cromwell’s prized flowers. In order to put an end to this she devised the simple trap be

Ivahew [28]

Answer:

The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.

Explanation:

Taking moment along point A

$\sum M_A=0\\F_{BC} sin 60 \times 48 -13 \times 8=0\\F_{BC}=2.502 lb_f$

Also normal stress is given as 100 psi

Now

$\sigma=\frac{F}{A}\\100=\frac{2.502}{\pi \frac{d^2}{4}}\\d^2 =\frac{4 \times 2.502}{100 \pi }\\d^2=0.031856\\d=\sqrt{0.031856}\\d=0.1785 inch$

The minimum diameter of the stick to bear the normal stress as 100 psi is 0.1785 in.

4 0

3 years ago