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3 years ago

15

a tuck at rest starts to move and accelerated by 4m/s in 25 secondd. what is the velocity of the truck at the end offf the time?

2 answers:

Reptile [31]3 years ago

8 0

Answer: 100 m/s
Explanation; 4x25

inysia [295]3 years ago

4 0

Answer:

100m/s

Explanation:

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If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w

DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

$d = 2\pi r$

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

$d = 2\pi r = 2\pi*6371 km = 40030.2 km$

Now, if we divide the above distance by the speed of the car we can find the time:

$t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d$

Therefore, Scobie will take 10 days to drive around Earth's equator.

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Hey can someone send me the answer to this

bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.

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3 years ago

What happens when two waves meet

olganol [36]

Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium. The interference of waves causes the medium to take on a shape that results from the net effect of the two individualwaves upon the particles of the medium.Hope this help!

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3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

4 years ago

Doe anyone get this

8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

2 m/s²
0.19 m/s²
9.25 m/ s²
0.04 m/s²
100.39 m/s²

4 0

3 years ago