All categories

3 years ago

What is the difference between a distance time graph and a speed time graph?

Physics

1 answer:

Feliz [49]3 years ago

3 0

Answer:

speed-time graph is the acceleration of the particle or straight lines acceleration is constant

a distance-time graph represents that a particle has constant speed.

Explanation:

You might be interested in

3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2

goblinko [34]

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

4 0

3 years ago

phase, color, and ductility are all examples of what type of property? chemical changeable physical external

svet-max [94.6K]

<span>Phase, color, and ductility are all examples of physical external properties</span>

3 0

3 years ago

We now have two tappers that are tapping the surface of the water together. They are tapping in sync (both are hitting the surfa

ollegr [7]

Answer:

The cork moves up and down.

Explanation:

The cork moves up and down because of the waves produce due to tapping the surface of water. The cork experience movement due to placed on the water surface so when the wave passes through the medium by tapping of water, movement of cork also occur. Cork has high volume and lower mass so it floats on the water surface and experience movement due to waves.

5 0

3 years ago

Compare the strengths of UV light and microwaves (in Hz). Which type of light is more powerful and how do you know?

TEA [102]

Answer:

UV light is more powerful as it has greater energy.

Explanation:

The energy propagated by electromagnetic waves ( light ) through vacuum or medium is known as electromagnetic radiation.

The frequency/wavelength range of electromagnetic radiation is known as electromagnetic spectrum. The electromagnetic spectrum ranging from gamma ray to radio waves.

Frequency range of UV light = ( 8 x 10¹⁴ to 3 x 10¹⁶ ) Hz

Frequency range of Microwaves = ( 300 x 10⁶ to 300 x 10⁹ ) Hz

Ratio of UV light to Microwaves = ( $\frac{8\times10^{14} }{300\times10^{6} }$ to $\frac{3\times10^{16} }{300\times10^{9} }$ )

= ( 2.66 x 10⁶ to 1 x 10⁸ )

Energy of electromagnetic radiation is given by the relation:

E = hν

Here h is plank's constant and ν is frequency.

UV light is more powerful than Microwaves as frequency of UV light is greater than frequency of microwaves. Thus, by the above equation, the energy of UV light is more than energy of Microwaves.

5 0

3 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago