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2 years ago

A visual display of data or information is a

1 answer:

4 0

A visual display of data or information is called a graph. There are many types of graphs. These can include pie graphs, bar graphs, and many more. Graphs are useful, because they show you visually data which is helpful to many. Hope this helped

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The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

2 years ago

State two applications of electrostatic

kaheart [24]

Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.

6 0

2 years ago

The counter is 1.5 m high, and the bowl has a mass of 0.5 kg. How much gravitational energy is stored in the bowl-earth system?

alina1380 [7]

Answer:

7.35 J

Im assuming, upon answering the question, that the gravity in this scenario is 9.8? As 9.8 is the gravitational force upon the earth.

5 0

2 years ago

Please help me people

saveliy_v [14]

Answer :

(1) The density of asphalt is, $1200kg/m^3$

(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.

(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

Explanation :

Part 1 :

As we are given:

Mass of block = 90 kg

Volume of block = $0.075m^3$

Formula used :

$\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}$

Now put all the given values in this formula, we get:

$\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3$

Thus, the density of asphalt is, $1200kg/m^3$

Part 2(a) :

As we are given that:

Length of aluminium sheet = 35 cm

Width of aluminium sheet = 11 dm

Thickness of aluminium sheet = 15 mm

Now we have to convert these dimensions into meters.

Conversions used:

1 cm = 0.01 m

1 dm = 0.1 m

1 mm = 0.001 m

Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m

Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m

Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m

Part 2(b) :

First we have to calculate the volume of aluminium sheet.

Volume of aluminum sheet (cuboid) = Length × Width × Thickness

Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m

Volume of aluminum sheet (cuboid) = 0.005775 m³

Now we have to calculate the mass of aluminium sheet.

$\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}$

$2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}$

$\text{Mass of aluminium}=15.59kg$

Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

7 0

2 years ago

Read 2 more answers

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

2 years ago