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2 years ago

Note: the 2 in the answers are exponents

Physics

1 answer:

stiks02 [169]2 years ago

4 0

Final speed after collision with the wall

$v_f = -0.4 m/s$

before collision the speed of ball initially

$v_i = 0.6 m/s$

time taken for the collision

$\Delta t = 0.2 s$

now as per the formula of acceleration we know that

$a = \frac{v_f - v_i}{\Delta t}$

now plug in all values in it

$a = \frac{-0.4 - 0.6}{0.2}$

$a = -5.0 m/s^2$

so acceleration is - 5 m/s/s for above situation

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Explain how electromagnetic waves are produced. I

madreJ [45]

Answer:

Electromagnetic are produced by gases and air and so much many things

Explanation:

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2 years ago

91.0x26x504 ...............

Hoochie [10]

Answer:

1192464

Explanation:

91.0 times 26 = 2366

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2 years ago

Read 2 more answers

If you went to a planet that had the twice the radius as Earth, but the same mass, a 1 kg pineapple would have a weight of

kicyunya [14]

Use the law of universal gravitation, which says the force of gravitation between two bodies of mass m₁ and m₂ a distance r apart is

F = G m₁ m₂ / r²

where G = 6.67 x 10⁻¹¹ N m²/kg².

The Earth has a radius of about 6371 km = 6.371 x 10⁶ m (large enough for a pineapple on the surface of the earth to have an effective distance from the center of the Earth to be equal to this radius), and a mass of about 5.97 x 10²⁴ kg, so the force of gravitation between the pineapple and the Earth is

F = (6.67 x 10⁻¹¹ N m²/kg²) (1 kg) (5.97 x 10²⁴ kg) / (6.371 x 10⁶ m)²

F ≈ 9.81 N

Notice that this is roughly equal to the weight of the pineapple on Earth, (1 kg)g, where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, so that [force of gravity] = [weight] on any given planet.

This means that on this new planet with twice the radius of Earth, the pineapple would have a weight of

F = G m₁ m₂ / (2r)² = 1/4 G m₁ m₂ / r²

i.e. 1/4 of the weight on Earth, which would be about 2.45 N.

7 0

2 years ago

The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids

BartSMP [9]

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

$100\times5\times60= m(1700)20+(0.63-m)(4200)20$

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

3 0

2 years ago

The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve

solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d = 0.107 mm = 0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 / 4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0

2 years ago