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3 years ago

Which SI unit is the correctly abbreviated for describing the mass on an object.

Physics

1 answer:

djverab [1.8K]3 years ago

4 0

S.I. Unit of mass is Kilogram which is denoted by Kg

In short, Your Answer would be Option C

Hope this helps!

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A car travels 48 km in 1.2 hours. What is the average speed of the car in km/hr

Alborosie

Answer:

average speed of car is 40 km/h

Explanation:

if car goes 48 km per 1.2 hours , the car goes 40 km per hour. so average speed of car is 40km/h

5 0

3 years ago

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws

cricket20 [7]

Answer:

Explanation:

Hope this helps!!!!

6 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

Which of the following statements is the best definition of climate?

noname [10]

Climate is a particular place's distance from the equator

4 0

3 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Marrrta [24]

Answer:

h = 2.64 meters

Explanation:

It is given that,

Mass of one ball, $m_1=3\ kg$

Speed of the first ball, $v_1=20\ m/s$ (upward)

Mass of the other ball, $m_2=2\ kg$

Speed of the other ball, $v_2=-12\ m/s$ (downward)

We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :

$V=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$