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Airida [17]
3 years ago
12

Question In order for work to be done, what three things are necessary

Physics
2 answers:
Taya2010 [7]3 years ago
8 0
You will need work ethic, self control, and a good mindset
Misha Larkins [42]3 years ago
5 0
Perseverance, good mind set, and work ethic
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An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?
Marizza181 [45]
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s 
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
6 0
3 years ago
Which image illustrates refraction?
maks197457 [2]

Answer:

B illustrates refraction

3 0
3 years ago
At 20 C, a steel rod of length 40.000 cm and a brass rod
balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

<u>Step 1:</u> <u>identify the given parameters and standard parameters</u>

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

                     stress in the rod =?

 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

--------------------------------------------------------------------------------------------

<u>Step 2:</u> <u>calculate the stress in each rod</u>

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

<u>Step 3:</u> <u>calculate the new length of each rod</u>

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0
2 years ago
Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula
denpristay [2]

Answer:

2

Explanation:

6 0
3 years ago
A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire
gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

V_T^2=v_N^2-v_W^2

V_T = \sqrt{v_N^2-v_W^2}

Travel north 2mph and west to 1mph, then,

V_T = \sqrt{2^2-1^2}

V_T = \sqrt{3}

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

v= \frac{\Delta x}{t}

t = \frac{\Delta x}{v}

t = \frac{2*(1mile)}{\sqrt{3}mph}

t = 1.15hour

Therefore the total travel time for the man is 1.15hour.

3 0
3 years ago
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