Vectors are arrows that tell us two things about force...WEIGHT and DIRECTION

A particle moves along the x-axis with velocity given by v(t)=3t2+6t for time tâ¥0. If the particle at position x=2 at time t=0,

sergey [27]

Answer:

b) 6

Explanation:

Given

v(t)=3t²+6t

X(0) = 2

X(1) = ?

Knowing that

v(t)=3t²+6t = dX/dt

⇒ ∫dX = ∫(3t²+6t)dt

⇒ X - X₀ = t³ + 3t²

⇒ X(t) = X₀ + t³ + 3t²

If X(0) = 2

⇒ X(0) = X₀ + (0)³ + 3(0)² = 2

⇒ X₀ = 2

then we have

X(t) = t³ + 3t² + 2

when

t = 1

X(1) = (1)³ + 3(1)² + 2

X(1) = 6

8 0

4 years ago

Read 2 more answers

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi

Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b) 195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

3 0

3 years ago

Determine the average distance between the Earth and the Sun. Then calculate the average speed of the Earth in its orbit in kilo

FrozenT [24]

Answer:

The average speed of the Earth in its orbit is 4.74 km/s.

Explanation:

Distance between the Earth and the Sun ,r= 149.60 million km = $149.60\times 10^6 km$

Considering the movement of Earth around the Sun follows circular path.

The circumference of circular path will be give by :

$D=2\pi r=2\times 3.14\times 149.60\times 10^6 km$

Time taken by Earth to complete revolution around earth ,T= 1 year

1 year = 365 days

1 year = 31,536,000 seconds

T = 31,536,000 seconds

Speed of the Earth around the Sun :S

$S=\frac{D}{T}=\frac{149.60\times 10^6 km}{31,536,000 s}$

$S=4.74 km/s$

The average speed of the Earth in its orbit is 4.74 km/s.

8 0

3 years ago

A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic

vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 $m^{2}$ </h3>

Explanation:

Given :

No. of turns in the first coil $N_{1} = 41$

No. of turns in the second coil $N_{2} = 123$

Area of first coil $A_{1} = 0.074 m^{2}$

According to the law of electromagnetic induction,

Induced emf = $-N \frac{d \phi}{dt}$

Where $\phi =$ magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

$-\frac{N_{1} d\phi _{1} }{dt_{1} } = -\frac{N_{2} d\phi _{2} }{dt_{2} }$

$\frac{N_{1} A_{1} dB_{1} }{dt_{1} } = \frac{N_{2} A_{2} dB_{2} }{dt_{2} }$

$A_{2} = \frac{N_{1} A_{1} }{N _{2} }$

$A_{2} = \frac{41 \times 0.074 }{123 }$

$A_{2} = 0.0246 = 0.025 m^{2}$

Therefore, the area of second coil is ≅ 0.025 $m^{2}$

4 0

4 years ago

Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the

stealth61 [152]

Answer:

$\frac{50}{\pi }$ Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф) -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t) ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100 [divide both sides by 2]

π f = 50

f = $\frac{50}{\pi }$ Hz

Therefore, the frequency of the voltage is $\frac{50}{\pi }$ Hz

7 0

3 years ago