A sharp knife cut things easily

If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long

vladimir2022 [97]

Answer:

The value is $t = 3.6 \ days$

Explanation:

From the question we are told that

The speed is $v = 450.0 km/h$

The radius of the earth is $R = 6200 \ km$

Generally the circumfernce of the earth is mathematically evaluated as

$C = 2\pi R$

=> $C = 2 * 3.142 * 6200$

=> $C = 38960.8 \ km$

Generally the time taken is mathematically represented as

$t = \frac{38960.8}{450.0}$

$t = 86.6 \ hr$

Converting to days

$t = \frac{86.6}{24}$

=> $t = 3.6 \ days$

7 0

3 years ago

A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t

zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

8 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom

amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

$v_1t + v_2t = d$

$4*t + 28*t = 3.2 km$

$t = \frac{3.2}{32} = 0.1 hr$

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect

4 0

3 years ago

What are four types of pathogens and what type of disease or sickness do they cause?

Zanzabum

Bacteria( cholera)

Virus( AIDS)

Fungi(athlete's foot)

Protist( malaria)

Pleasssssssse mark me as brainliest

7 0

3 years ago

A sharp knife cut things easily​

A sharp knife cut things easily