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Alexxx [7]
3 years ago
14

I ship travels north at 15 km an hour for five hours what is the ships this placement in kilometers

Physics
1 answer:
andrew11 [14]3 years ago
8 0
What you would is is multiply 15km by 5 and end up with 75km
You might be interested in
what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3
Oduvanchick [21]
V=m×d
m=v/d
m=113/21.4
m=5.28g


6 0
3 years ago
Read 2 more answers
Atoms are the smallest portions of an element that retain the original characteristics of the element. True or false
kozerog [31]

Answer:

True

Explanation:

Going even smaller than atoms would get you to subatomic particles such as quarks. From there, it is impossible to distinguish elements. So, yes, atoms are the smallest portions of an element that retains the original characteristic of the element.

4 0
3 years ago
PLZZZ HELPPP ASAP<br> I really need help as soon as possible
julsineya [31]

Answer:

Friction

Explanation:

As the toy cars rolls away, more friction is created. The more friction there is, the more friction on surface rubs against another which creates friction which in-term slows it down. Hope this helps.

4 0
2 years ago
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0
3 years ago
Read 2 more answers
You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters
olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car m=2400 kg

Speed of car u=14 m/s

Distance traveled before coming to halt s=23.2 m

Let \muthe coefficient of friction

Maximum deceleration road can provide during motion is

a=\mu g

using v^2-u^2=2 as

0-14^2=2\cdot (-\mu g)\cdot 23.2

\mu =\frac{196}{454.72}

\mu =0.431

7 0
3 years ago
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