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jasenka [17]
2 years ago
6

What is conserved in a thermodynamically closed system?

Physics
1 answer:
kherson [118]2 years ago
3 0

Explanation:

C. Both energy and matter

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Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second
tiny-mole [99]

Answer:

–1.1 × 10^11 N

Explanation:

I did the assignment

4 0
2 years ago
how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5
Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

4 0
3 years ago
Read 2 more answers
A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
Marat540 [252]

Answer:

a)  19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

b)

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

5 0
3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
2 years ago
Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow
bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

F=\frac{\Delta p}{\Delta t}

where

F is the force exerted

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be rewritten as

\Delta p = m(v-u)

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

F=\frac{m(v-u)}{\Delta t}

In this problem we have:

\frac{m}{\Delta t}=0.030 kg/s is the mass rate

u=-15 m/s is the initial velocity

v=+15 m/s is the final velocity

Therefore, the force exerted by the hail on the roof is:

F=(0.030)(+15-(-15))=0.9 N

6 0
3 years ago
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