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valentinak56 [21]
3 years ago
8

If a truck travels 100 miles in 2 hours, what is its speed?

Physics
2 answers:
densk [106]3 years ago
6 0

Answer:

50 miles/hr

Explanation:

Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour

miskamm [114]3 years ago
5 0

Answer:50  miles per hour 50/1hr

Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1

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zaharov [31]
D - tertiary consumer

This is because it is the farther up to food chain.  
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Write the dimension of a / b in the x = at + bt2. Where x is the distance and t is the time?​
Luba_88 [7]

The dimension of a/b where x is the distance and t is the time is T

Given the expression

x = at + bt²

where

x is the distance

t is the time

Based on the homogeneity principle, the expression on the left-hand side must be equal to that on the right. Hence;

x = at

a = \frac{x}{t}

Since x is the distance and distance is measured in metres, the dimension equivalent will be the length 'L'

Since t is the time and time is measured in seconds, the dimension equivalent will be the seconds 'T'

a=\frac{L}{T}

Similarly;

x  = bt²

b=\frac{x}{t^2}\\b=\frac{L}{T^2}

Next is to get a/b;

\frac{a}{b} =  \frac{L}{T} \div \frac{L}{T^2}\\\frac{a}{b} = \frac{L}{T}*\frac{T^2}{L}  \\\frac{a}{b} =\frac{T^2}{T}\\\frac{a}{b} =T

Hence the dimension of a/b is T

4 0
3 years ago
¿Diferencias entre movimiento vertical y horizontal?​
Ronch [10]

Answer:

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5 0
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Read 2 more answers
a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance
Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

v=u+at

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

t = 5 s is the time

So, the final velocity is

v=0+(2)(5)=10 m/s

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

v^2-u^2=2as

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

v = 10 m/s is the final velocity

Solving for s,

s=\frac{10^2-0^2}{2(2)}=25 m

3 0
3 years ago
The acceleration of a vertically thrown ball at the top of its path is?
ser-zykov [4K]
C between zero and g
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