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4 years ago

Express 0.0097 in scientific notation

Physics

1 answer:

Maru [420]4 years ago

3 0

Explanation:

Move the decimal point until it is behind the first non-zero digit.

In this case, we move it 3 places to the right so it is behind the 9.

Therefore, 0.0097 = 9.7×10⁻³.

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What is the momentum of an object that has a mass of 0.03 grams and a velocity of 1200 m/s2

ruslelena [56]

answer is 36

because the formulae of momentum is

mass×velocity

3 0

3 years ago

A mass attached to a spring oscillates with a period of 22 sec. After 22 kg are added, the period becomes 33 sec. Assuming that

polet [3.4K]

Answer:

Mass of 17.854 kg is only attached to the spring

Explanation:

We have given time period in first case is 22 sec

Let initially mass is m

After 22 kg are added the period becomes 33 sec

Time period of spring mass system is

$T=2\pi \sqrt{\frac{m}{k}}$ , here m is mass and k is spring constant

From the relation we can see that

$\frac{T_1}{T_2}=\sqrt{\frac{m}{m+22}}$

$\frac{22}{33}=\sqrt{\frac{m}{m+22}}$

Squaring both side

$0.444={\frac{m}{m+22}}$

$0.444m+9.777=m$

m = 17.584 kg

So mass of 17.854 kg is only attached to the spring

7 0

3 years ago

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc

nasty-shy [4]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

4 0

3 years ago

Two narrow slits 84 mum apart are illuminated with light of wavelength 650 nm. What is the angle of the m = 3 bright fringe in r

Digiron [165]

Answer:

The angle is 1.33°.

Explanation:

Given that,

Distance $d=84\times10^{-6}\ m$

Wavelength = 650 nm

Number of fringe = 3

We need to calculate the angle

Using formula of angle for brightness

$d\sin\theta=m\lambda$

$\theta=\sin^{-1}\dfrac{m\lambda}{d}$

Where, d = distance

$\lambda$ =wavelength

m = number of fringe

Put the value into the formula

$\theta=\sin^{-1}\dfrac{3\times650\times10^{-9}}{84\times10^{-6}}$

$\theta=1.33^{\circ}$

Hence, The angle is 1.33°.

3 0

4 years ago

A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot

FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

$m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0$

with m = 1 slug and k = 9 lb/ft, the equation becomes

$\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0$

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ where k is an integer

3t = ±π/6 + 2kπ - 4π/3

t = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0

3 years ago