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2 years ago

Write a hypothesis about the use of an object's physical

Physics

1 answer:

KengaRu [80]2 years ago

3 0

If the mass of the object and the volume of the object is determined;

Then, the density of the object is determined by taking the ratio of the mass and volume.

<h3>What is density of an object?</h3>

The density of an object is the ratio of the mass and volume of that object.

Mathematically;

Density = mass/volume

To determine the density of an object therefore, the physical characteristics of mass and the volume of the object are measured.

The mass of the object is obtained using a scale or a balance.

The volume of the object if a solid is obtained using a displacement bottle. If it is a liquid, a measuring cylinder is used.

The density of the object is then obtained by taking the ratio of the mass and the volume of the object.

In conclusion, the density of an object is determined from the volume and mass ratio.

Learn more about density at: brainly.com/question/1354972

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Which statement is true about voice?

Ksju [112]

The answer is A. voice uses a wider range of pitch and volume as compared to speaking

3 0

4 years ago

Read 2 more answers

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

Svet_ta [14]

Answer:

The speed is $\sqrt{2}v_{0}$ .

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$ ....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$ , then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$ .

6 0

3 years ago

The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows

Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000

= 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = $P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W$

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

$\eta =\frac{Power output}{Power input}$

$\eta =\frac{716}{83475}=0.858$

Thus, the efficiency is 85.8 %.

7 0

3 years ago

A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is fo

Alex787 [66]

Answer:

ρ = 1.6*10⁻⁸ Ω/m.

Explanation:

Applying Ohm's Law to the wire, assuming that it can be treated as a pure resistance, the resistance of the wire can be obtained as follows:

$R = \frac{V}{I} = \frac{9.11V}{36.0A} = 0.253 \Omega (1)$

At the same time, we know that there exists a relationship between the resistance, the resistivity ρ, the length L and the area A of the wire, that is given for the following expression:

$R = \rho* \frac{L}{A} (2)$

The area of the circular section of the wire, can be expressed as a function of the diameter d, as follows:

$A = \frac{\pi*d^{2} }{4} = \frac{\pi*(0.002m)^{2}}{4} = \pi*10e-6 (3)$

Replacing the left side of (2) by (1), and (3) on the right side, we can solve for the resistivity ρ as follows:

$\rho = \frac{R*A}{L} = \frac{0.253\Omega*\pi*10e-6}{50.0m} = 1.6e-8 \Omega/m$

ρ = 1.6*10⁻⁸ Ω/m

4 0

3 years ago

We would like to use the relation V(t)=I(t)RV(t)=I(t)R to find the voltage and current in the circuit as functions of time. To d

drek231 [11]

Answer:

V = -RC (dV/dt)

Solving the differential equation,

V(t) = V₀ e⁻ᵏᵗ

where k = RC

Explanation:

V(t) = I(t) × R

The Current through the capacitor is given as the time rate of change of charge on the capacitor.

I(t) = -dQ/dt

But, the charge on a capacitor is given as

Q = CV

(dQ/dt) = (d/dt) (CV)

Since C is constant,

(dQ/dt) = (CdV/dt)

V(t) = I(t) × R

V(t) = -(CdV/dt) × R

V = -RC (dV/dt)

(dV/dt) = -(RC/V)

(dV/V) = -RC dt

∫ (dV/V) = ∫ -RC dt

Let k = RC

∫ (dV/V) = ∫ -k dt

Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.

In V - In V₀ = -kt

In(V/V₀) = - kt

(V/V₀) = e⁻ᵏᵗ

V = V₀ e⁻ᵏᵗ

V(t) = V₀ e⁻ᵏᵗ

Hope this Helps!!!

5 0

4 years ago