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2 years ago

If the specific gravity of copper is 8.91, it weighs:

Physics

1 answer:

denis-greek [22]2 years ago

6 0

Answer:

Option D is the correct answer.

Explanation:

The specific gravity of copper is 8.91.

We have

$\frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3$

We also have

1 kg = 2.205 lb

1 m = 3.28 ft

Substituting

$\texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3$

Option D is the correct answer.

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Hey help me plzzzzz i will mark brainliest

vodomira [7]

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

7 0

2 years ago

Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in

USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

$W=F\cdot d$

Then, we can derive the magnitude of the force as:

$F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N$

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

$F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N$

6 0

2 years ago

What type of wave does not need matter to carry energy?

marishachu [46]

Non- mechanical wave does not need matter to carry energy.

e.g:- Light

3 0

3 years ago

Read 2 more answers

An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A

melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

$\omega = \frac{\theta}{t}$

Where θ: is the angular displacement = π (since it moves halfway through a complete circle)

We have:

$t = \frac{\theta}{\omega} = \frac{\theta}{v/r}$

Where:

v: is the tangential speed

r: is the radius

The radius can be found equaling the magnetic force with the centripetal force:

$qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}$

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

$t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s$

Therefore, the time that takes for an alpha particle to move halfway through a complete circle is 4.21x10⁻⁷ s.

I hope it helps you!

4 0

3 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago