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3 years ago

The equation for distance is d = st. If a car has a speed of 85 m/s and travels

Physics

1 answer:

Ludmilka [50]3 years ago

4 0

Answer:

D) 10 200 m

Explanation:

Speed is known as the rate of change of distance.

$speed=\frac{distance}{time}$

distance = speed * time

= 85 ms⁻¹ * 120 s

= 10200 m

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Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens

Nutka1998 [239]

Answer:

1) Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

1) Radius of the grain,r = 64.8 µm = $6.48\times 10^{-5}m$

Volume of the sphere = $\frac{4}{3}\pi r^3$

Volume of the grain of a sand:

$V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3$

$V=1.1397\times 10^{-12} m^3$

Density of a grain of sand = $d=2600 kg/m^3$

Mass of a grain of a sand = M

$d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}$

$M=2.9632\times 10^{-9} kg$

Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) Surface are of sphere: $4\pi r^2$

Surface area of a grain:

$A=4\times 3.14\times (6.48\times 10^{-5}m)^2$

$A=5.2766\times 10^{-8} m^2$

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= $6a^2$

$A'=6\times (0.514 m)^2=1.5851 m^2$

let the number grains with area equal to total surface area of cube be x.

$A'=A\times x$

$x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7$

Volume of x number of grains :V'

$V'=V\times x$

$V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7$

$V'=3.4236\times 10^{-5} m^3$

Mass of $3.4236\times 10^{-5} m^3$ of sandL:

= $3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg$

0.08901 kg of sand would have surface area equal the surface area of the cube.

8 0

3 years ago

A right triangle with the base labeled 40 meters and the height labeled 30 meters. The hypotenuse is a dotted arrow labeled R. W

Pepsi [2]

Answer:

The magnitude of the resultant vector R is 50 meters ⇒ 2nd answer

Explanation:

The resultant vector is the vector sum of two or more vectors

If the two vectors perpendicular to each other, then the magnitude of

the resultant vector is the square root of the sum of their squares

If x and y are two vectors perpendicular to each other, then the

magnitude of its resultant vector R is:

→ $R=\sqrt{x^{2}+y^{2}}$

Lets solve the problem

A right triangle with the base labeled 40 meters and the height labeled

30 meters

The hypotenuse is a dotted arrow labeled R

→ The base and the height of the right triangle are perpendicular

→ The hypotenuse is the resultant vector of them

Assume that x represents the base of the triangle and y represents the

height of it

By using the rule above

→ x = 40 m , y = 30 m

→ $R=\sqrt{x^{2}+y^{2}}$

→ $R=\sqrt{(40)^{2}+(30)^{2}}$

→ $R=\sqrt{1600+900}$

→ $R=\sqrt{2500}=50$

The magnitude of the resultant vector R is 50 meters

7 0

3 years ago

5N of force is applied to move a large nail a distance of 10 cm from an electromagnet on a frictionless table. The nail is then

DedPeter [7]

Answer:

C) The rate of change from potential to kinetic energy is exponential.

Explanation:

As we know that total mechanical energy must be conserved here

As we know that there is no friction force on this system of electromagnet and nail

So here we can say that that

Magnetic potential energy of the nail + electromagnet system will convert into kinetic energy of the nail as it is released.

So we will have

Initial potential energy = work done to move it away by 10 cm

$U = 5(0.10)$

$U = 0.5 J$

now as the nail is released then this potential energy will start to convert into kinetic energy

So here correct answer must be

C) The rate of change from potential to kinetic energy is exponential.

7 0

3 years ago

Name three common minerals in silicates

rjkz [21]

Quartz, feldspa, mica, amphibole, pyroxene and olivine

3 0

3 years ago

A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

thus where the pipe is horizontal we have

z₁ = z₂ hence the above equattion becomes

$P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}$

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

5 0

4 years ago