What is the tendency of two masses alone in the universe to drift together a result of?

Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height

diamong [38]

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

5 0

3 years ago

Freeeee pointsssssss. i hope yall have a good as well!! Also, whats something you regret doing most?

leonid [27]

Answer:POGGERS thx i rl needed this have a good day also something a regret the most is searching up henhai i am now addicted

6 0

3 years ago

Read 2 more answers

Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it

makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

Here, we have been given the height through which the ball drops;

Height of drop = 14.5m - 1.9m = 12.6m

The right motion equation is;

V² = U² + 2gh

V is the final velocity

U is the initial velocity = 0

g is the acceleration due to gravity = 9.8m/s²

h is the height

Now insert the parameters and solve;

V² = 0² + 2 x 9.8 x 12.6

V² = 246.96

V = √246.96 = 15.7m/s

4 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

một ô tô chuyển động nhanh dần đều.Sau 10s vận tốc của xe tăng từ 2m/s lên 6m/s.Tính gia tốc của ô tô

Serjik [45]

Answer:

áp dụng công thức í, mình thấy câu này có rắc rối gì đâu

8 0

3 years ago