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Veronika [31]
3 years ago
11

Which of these is an organism??

Physics
1 answer:
elena55 [62]3 years ago
5 0
You need to add a picture or answers!
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Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
nignag [31]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

5 0
3 years ago
All objects are in motion relative to the sun. <br> a. True<br> b. False
Gre4nikov [31]
That depends on what "objects" refer to.
If "objects" refer to the ones on Earth, then it is TRUE. These objects are in motion relative to the sun same as with Earth itself.

<span>If "objects" refer to the ones outside earth, then it may be a TRUE or FALSE depending on how far or near they are from the sun.</span>
6 0
3 years ago
4. A little boy pushes a wagon with his dog in it. The mass of the dog and
Artyom0805 [142]

Answer:

<h2>59.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 70 × 0.85

We have the final answer as

<h3>59.5 N</h3>

Hope this helps you

6 0
3 years ago
Are the items in the list renewable or nonrenewable resources?
Elza [17]
N
R
R
R
N
Hope this helps.
8 0
3 years ago
Read 2 more answers
An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t
ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out a_{x} and time t.

As we know that the initial speed is zero. So

(a)  

v_{0x} = 0

x - x_{o} = 1.25×10^{-2}m

v_{x} = 3.3×10^{6}m/s

v^{2} _{x} =  v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )

a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }

   = \frac{(3.3 * 10^{6})^{2}  - 0 }{2(1.25 * 10^{-2}) }

   = 4.356×10^{14} m/s²

(b)

v_{x} = v_{ox} + a_{x}t

t = v_{x} - vo_{x}/a_{x}

t = \frac{3.00 * 10^{6} }{4.356*10^{14} } = 6.8870×10^{-9}s

(c)

ΣF_{x} = ma_{x}

       = (9.11×10^{-31})(4.356×10^{14}m/s²)

       = 3.968×10^{-16} N

6 0
3 years ago
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