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5 months ago

15

starting at a position of 0 m, if the man is moving at a constant velocity of 2 m/s, it will take seconds for him to reach a pos

ition of 12 m.

1 answer:

Harman [31]5 months ago

4 0

The calculated time is 6 seconds.

Time is defined by physicists as the flow of events from the past through the present and into the future. In essence, a system is timeless if it is unchanging. When describing events that take place in three-dimensional space, time can be thought of as the fourth dimension of reality. Even if time isn't directly connected to energy, it is undoubtedly connected to many other components of energy. For instance, the movement of energy across the universe can affect the direction of time (from the past to the future).

V= 2 m/s

d=12 m

t=?

we know that velocity=displacement / time

time= displacement / velocity

= 12/2

=6 seconds

the calculated time is 6 seconds.

Learn more about time here-

brainly.com/question/28050940

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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

2 years ago

15. Some people have observed that animal behavior changes just before an earthquake. What might explain this?

Ber [7]

Answer:

15: Animals are able to detect the first of an earthquake's seismic waves—the P-wave, or pressure wave, that arrives in advance of the S-wave, or secondary, shaking wave. This likely explains why animals have been seen snapping to attention, acting confused or running right before the ground starts to shake.

16: No, it is not even remotely true. Microwaves cook food from the outside in, just like a regular oven. In fact, most of the cooking on the inside of the food, depending on its thickness, is done by heat conduction from the outside surfaces inwards, as the microwaves do not actually penetrate that far into the food.

3 0

1 year ago

How much pure gold (24/24) is to be added to 15.0 grams of 14 K gold (14/24) to obtain 18 K gold (18/24)?

Arlecino [84]

Answer:

10 gram gold is added

Explanation:

given data

pure gold = (24/24)

added = 15 grams

gold = 14 K ( 14/24)

gold = 18 K ( 18/24)

to find out

How much pure gold added

solution

we know here that when we add gold to get 20 K gold or 22K

so we added here 15 gram 14 K

we consider here m is pure gold added

so by composition here

we get

15 (14K) + m ( 24 K) = ( 15 + m ) (18) ...................1

solve it and find m

m = 10

so 10 gram gold is added

8 0

2 years ago

What is the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20 m/s?

Lesechka [4]

Ke=1/2mv^2
Ke= 0.5*30*20^2
Ke= 6000joules

8 0

2 years ago

A piece of aluminum (bulk modulus 7.1 x 1010 N/m2) is placed in a vacuum chamber where the air pressure is 0.781 x 105 Pa. The v

Jet001 [13]

Answer:

${\dfrac{\Delta V}{V}}=11\times 10^{-7}$

Explanation:

Given that

Bulk modulus ,K = 7.1 x 10¹⁰ Pa

The pressure P₁ = 0.781 x 10⁵ Pa

The final pressure P ₂ = 0 Pa

We know that bulk modulus given as

$K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

Now by putting the values

$K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}$

$7.1\times 10^{10}=\dfrac{0.781\times 10^5}{\dfrac{\Delta V}{V}}$

${\dfrac{\Delta V}{V}}=\dfrac{0.781\times 10^5}{7.1\times 10^{10}}$

${\dfrac{\Delta V}{V}}=11\times 10^{-7}$

Therefore fractional change will be 11 x 10⁻⁷.

3 0

2 years ago