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2 years ago

Select the correct answer.

Physics

2 answers:

Alika [10]2 years ago

6 0

Answer:

B :)

Explanation:

r-ruslan [8.4K]2 years ago

3 0

Answer:

That would be B. Hope this helps!

Explanation:

Orion's Belt or the Belt of Orion, also known as the Three Kings or Three Sisters, is an asterism in the constellation Orion. It consists of the three bright stars Alnitak, Alnilam and Mintaka. Looking for Orion's Belt in the night sky is the easiest way to locate Orion in the sky.

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Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

8 0

2 years ago

A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is

Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

= √(T/A×ρ)

= √[2.96×10^4/(4.49×10^-3×7860)]

= 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0

2 years ago

Some sharks can swim at average cruising speeds of three miles per hour. If a shark swam at that average speed for seven hours,

kari74 [83]

Answer:

21 miles

Explanation:

3 miles an hour for 7 hours

Its simply 7m*3m/hr=21 miles

8 0

2 years ago

The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70

MrRa [10]

Answer:

a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

C = $\epsilon_o \frac{A}{d}$

C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

C = $\frac{Q}{\Delta V}$

Q₀ = C ΔV

Q₀ = 2.62 10⁻¹² 8.70

Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

C₁ = 1.04 10⁻¹² F

C₁ = Q₀ / ΔV₁

ΔV₁ = Q₀ / C₁

ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

ΔV₁ = 21.9 V

b) initial stored energy

U₀ = $\frac{Q_o}{ 2C}$

U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)

U₀ = 99.2 10⁻¹² J

c) final stored energy

U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)

U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

W = U_f - U₀

W = (249.2 - 99.2) 10⁻¹²

W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0

2 years ago