What is the primary function of the small intestines

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

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Answer:

(a)0.0675 J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago

Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply

zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0

3 years ago

How many kindergarteners do you think you can take on a fight before getting tired or over powered?

yulyashka [42]

Is this a serious question ?

4 0

3 years ago

Read 2 more answers

Write the nuclear equation for the release of the beta particles by Pb-210.

Liula [17]

210 Pb ---> -ie + 210 B:
84 8.3

3 0

4 years ago

Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 25.0 m from each other. Bein

Lerok [7]

Answer:

A) 199.78 J

B) 9.292x10^14 J

C) 4.2x10^7 m/s

D) 0.65 m

E) 1.13x10^-8 sec

D) 2.94x10^-9 sec

Explanation:

mass of ball = 0.0580 kg

A)

If smashed at v = 83.0 m/s, KE is

KE = 0.5mv^2

= 0.5 x 0.0580 x 83.0^2

= 199.78 J

B) if returned at v = 1.79×10^8 m/s, KE will be

KE = 0.5mv^2

= 0.5 x 0.0580 x (1.79×10^8)^2

= 9.292x10^14 J

C) during Einstein's return, velocity of rabbit relative to players is

Vr = 2.21×108 m/s

Rabbit's velocity relative to ball = 2.21×10^8 - 1.79×10^8

= 4.2x10^7 m/s

D) the rabbit's speed approaches the speed of light so we consider relativistic effect. The rabbit's measured distance is

l = l°( 1 - v^2/c^2)

= 2.5(1 - 2.21/3)

= 2.5 x 0.26

= 0.65 m

E) according to the players, the time taken by the rabbit is

t = d/v = 2.5/ 2.21×10^8

= 1.13x10^-8 sec

F) the time for rabbit as measured by rabbit is relativistic

t = t°( 1 - v^2/c^2)

= 1.13x10^-8 (1 - 2.21/3)

= 1.13x10^-8 x 0.26

= 2.94x10^-9 sec

7 0

4 years ago