Ask question

Ask question

All categories

3 years ago

5

A 4.7 kg solid sphere made of metal, whose density is 4000 kg /m3, is suspended by a cord. The density of water is 1000 kg/m3.Wh

en the sphere is immersed in water, the tension in the cord is closest to:

1 answer:

Vlad1618 [11]3 years ago

8 0

Answer:

34.58 N

Explanation:

From the question,

Tension in the cord = Weight of the solid metal- Upthrust

T = W-U................... Equation 1

But,

W = mg

Where m = mass of solid, g = acceleration due to gravity.

Given: m = 4.7 kg, g = 9.8 m/s²

W = 4.7(9.8) = 46.1 N.

U = (Density of water×Volume of water displaced.)gravity

U = D×V×g.

But,

V = Mass of solid/density of solid

V = 4.7/4000

V = 1.175×10⁻³ m³.

Therefore,

U = 1.175×10⁻³(1000)(9.8)

U = 11.52 N

Therefore,

T = 46.1-11.52

T = 34.58 N

You might be interested in

A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.

vekshin1

Answer:

A) If the paintball stops completely the magnitude of the change in the paintball’s momentum is $p=0.273kg*m/s$

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is $p=0.546kg*m/s$

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum $p=mv$ , so we have $p=0.0032kg*85.3m/s=0.273kg*m/s$

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

$p=2*0.0032kg*85.3m/s=0.546kg*m/s$ .

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

$F=\frac{\Delta p}{\Delta t}$ , as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0

3 years ago

** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

$V_p = 350,000 Volts$

$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

6 0

3 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

3 years ago

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

dedylja [7]

Answer:

$2.2\times 10^{-5} s$

Explanation:

We are given that

The wavelength of sound wave= $\lambda=3.3 cm=3.3\times 10^{-2}m/s$

1 cm/s= $10^{-2}m/s$

Speed of sound wave,v= $1522 m/s$

We have to find the period of the wave.

We know that

Frequency= $\nu=\frac{v}{\lambda}$

Using the formula

Frequency = $\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4}$ Hz

Time period= $\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s$

Using identity: $\frac{a^x}{a^y}=a^{x-y}$

Hence, the time period of the wave= $2.2\times 10^{-5} s$

4 0

3 years ago

The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force that

Furkat [3]

Answer:

Earth pulls the sun towards itself with a force equal to the ratio of the mass of the sun to the mass of Earth

5 0

3 years ago