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3 years ago

Jake drove 160 kilometres in 2 hours. What was his speed, in kmh-1

Physics

2 answers:

sdas [7]3 years ago

7 0

Answer:

80kmph

Explanation:

speed = d/t

d= distance

t= time

s=160/2

s=80kmph

ruslelena [56]3 years ago

5 0

Answer:

80 kmh

Explanation:

IDK lol i just divided it by 2 because he drove 80 kilometres in one hour

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If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. It i

Kaylis [27]

Answer:

a.0.120mm

b.1.58s

c.0.6329Hz

Explanation:

a. Given that 0.120mm is displaced from equilibrium, 0.120mm after 0.790s on opposite side:

-The amplitude is the maximum displacement from equilibrium.

-The object goes from x=+A to x=-A and back during one cycle.

#Hence, the amplitude of the motion is 0.120mm

b.Motion from maximum positive displacement to maximum negative displacement takes places during half the period of Simple Harmonic Motion(SHM)

$0.790s=T/2\\\\T=0.790s\times 2\\\\T=1.58s$

#Hence, the period of the motion is 1.58s

c. Frequency is calculated as one divided by the period of the motion.

From b above we know that the motions period is 1.58s

Therefore:

Frequency=1/period=1/1.58=0.6329Hz

#The frequency of the motion is 0.6329Hz

3 0

3 years ago

An astronaut orbits planet Y in her spaceship. To remain in orbit at 410km above the planet's center, she maintains a speed of 6

choli [55]

Formula for orbital speed, v = √(GM/R)

Where G is the universal gravitational constant, M = Central Mass,

R = Distance between centers of Mass.

Given. v = 68 m/s, M = ? , R = 410 km = 410000 m., G = 6.674 * 10⁻¹¹ Nm²/kg²

68 = √(GM/R)

68 = √(6.674 * 10⁻¹¹ * M/410000)

68² = (6.674 * 10⁻¹¹ * M)/410000

(68² * 410000) / 6.674 * 10⁻¹¹ = M

2.84 × 10¹⁹ = M

Mass of Planet Y = 2.84 × 10¹⁹ kg

3 0

3 years ago

A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that

ExtremeBDS [4]

That isn"t the right answer the correct answer is B.

4 0

3 years ago

Read 2 more answers

You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving dire

charle [14.2K]

Answer:Velocity = 6.325m/s

Directional angle= 18.43°

Explanation:

Using Right angle triangle

Let Velocity of ballon&hawk be VHB represent the height of the triangle.

Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.

Let Velocity of hawk and ground BE VHG represent the hypothesis.

Theta = opp/Adj= VHB/VBG

using pythagorean

VHG= SQRT(VHB^2+VBG^2)

VHG= sqrt(2^2+6^2)

VHG= sqrt(4+36)

VHG= 6.325m/s

Tan theta= 2/6

Tan theta =0.3333

Tan^-1 0.3333=18.43°

8 0

3 years ago

A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b

Lesechka [4]

Answer:

a) Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F , V = 3.4375 V ,

c) U = 54.7 nJ , d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

C = Q / V

Q = C V

can also be calculated using geometry consideration

C = e or A / d

we reduce to the SI system

A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

d = 1.53 cm = 1.53 10⁻² m

we substitute

Q = eo A / d V

Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

Q = k Q₀

Q = 80 397.57

Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

C = k C₀

C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

C = 1.157 10⁻¹⁰ F

V = Vo / k

V = 275/80

V = 3.4375 V

c) The stored energy is

U = ½ C V²

U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275²

U = 5.47 10⁻⁸ J

let's reduce to nJ

109 nJ = 1 J

U = 54.7 nJ

d) energy after submerging

U = ½ (kCo) (Vo / k) 2

U = ½ Co Vo2 / k

U = U₀ / k

U = 54.7 / 80 nJ

U = 0.68375 nJ

the energy change is

ΔU = U₀ -U

ΔU = 54.7 - 0.687375

6 0

3 years ago