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Digiron [165]
2 years ago
10

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

1°C. How much energy was transferred to the water? (Use a value of 9.8 m/s2 for g.)
Engineering
1 answer:
Orlov [11]2 years ago
6 0

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

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It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)
ycow [4]

Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

for example ; bronze and brass

so from above discussion it is clear that option (d) will be the correct option

8 0
3 years ago
Read 2 more answers
A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

   = 270 / 0.0144

δ  = 18750 lb/in² = 19,000 Psi

<u>Part (B)</u> The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

<u>Part (C)</u> The modulus of elasticity of the steel

E = δ/σ

   = 19,000 / 0.00063

E = 30,000,000 Psi

4 0
3 years ago
A cruise missile under test is moving horizontally at Ma =2 in the atmosphere at an elevation of 2000 m (Air temperature is 2 °C
telo118 [61]

Answer: the half-angle "alpha" of the Mach cone = 30⁰

Explanation:

To calculate the  half-angle "alpha" of the Mach cone.

we say ;

Sin∝ = 1 / Ma

given that Ma = 2

now we substitute

Sin∝ = 1 / 2

Sin∝ = 0.5

∝ = Sin⁻¹ 0.5

∝ = 30⁰

Therefore, the half-angle "alpha" of the Mach cone is 30⁰

3 0
3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
2 years ago
Which of the following are true about the American Wire Gauge?
harina [27]

Answer:

A. smallest wire is No. 12

7 0
2 years ago
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