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arlik [135]
3 years ago
12

330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass i

s 840 grams and initial temperature 29°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water?
Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw × Cw×(100 –T) = Ma × Ca × (T–29)

330×4.2×(100– T) = 890×0.9×(T–29)

1386(100 – T) = 801(T –29)

1386/801(100 – T) = T – 29

1.73(100 – T) = T – 29

173 –1.73T = T –29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C

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14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.
sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

3 0
3 years ago
An Earth satellite moves in a circular orbit 680 km above Earth's surface with a period of 98.15 min. What are (a) the speed and
kotykmax [81]

The solution is in the attachment

6 0
2 years ago
Can anyone help me with this question please​
JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

  1. v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0
3 years ago
If the numbers on the plate are 6.0 cm apart, and the spy satellite is at an altitude of 160 km , what must be the diameter of t
Alexxx [7]

Answer:

The diameter of the camera aperture must be greater than or equal to 1.49m

Explanation:

Let the distance separating two objects, x = 6.0 cm = 0.06 m

The distance between the observer and the two objects, d = 160 km = 160000 m

Let ∅ = minimum angular separation between the two objects that the satellite can resolve

tan( ∅) = x/d

Since there is minimum angular separation, tan( ∅) ≈∅

∅ = x/d

∅ = 0.06/160000

∅ = 3.75 * 10⁻⁷rad

For the satellite to be able to resolve the objects,

D ≥ 1.22λ/∅

λ = 560 nm = 560 * 10⁻⁹

D  ≥ 1.22 *  (560 * 10⁻⁹)/(3.75 * 10⁻⁷)

D  ≥ 149.33 * 10⁻² m

D  ≥ 1.49 m

7 0
2 years ago
An object thrown vertically upward with a speed of 6 m/s from a height of h meters takes 10 seconds to reach the ground. set up
nexus9112 [7]
430m. We use the displacement formula x=v*t+1/2*a*t^2. Set upward to be the positive direction, so v=6m/s, a=-g=-9.8m/s^2 (gravity is pulling the object down), and t=10s. The displacement is -h(it's negative because the ending point is the ground, which is negative relative to the starting point). -h=60-4.9*100, h=430 m. 
3 0
3 years ago
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