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Evgen
7 days ago
15

If you have an effusive personality, then it __________ of you. motor oil will __________ across the gravel if it spills out of

the can.
Physics
1 answer:
Keith_Richards [1K]7 days ago
6 0
The first word in the blank is "pours out of you," while the second word to fill in is "diffuse." This implies that a person with an effusive nature tends to openly express their feelings, just as spilled motor oil will spread out over gravel.
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Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
Softa [913]
The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s
7 0
13 days ago
Read 2 more answers
A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo
Ostrovityanka [942]

Response:

216000 W or 216 kW

Details:

Power: This refers to the rate at which energy is utilized or consumed. The standard unit for power is the Watt (W).

In general,

Power = Energy/time

P = E/t........................ Equation 1.

However,

E = 1/2mv²..................... Equation 2

with m representing mass, and v indicating velocity.

Substituting equation 2 into equation 1,

P = 1/2mv²/t...................... Equation 3

Assuming flow rate (Q) = m/t,

Q = m/t................ Equation 4

Insert equation 4 into equation 3,

P = Qv²/2........................ Equation 5

Where Q stands for flow rate, v is velocity, and P denotes power.

Given: Q = 120 kg/s, v = 60 m/s

Plug these values into equation 5,

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus, the potential power generation of the water jet is 216000 W or 216 kW.

5 0
15 days ago
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [987]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
4 days ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Keith_Richards [1034]
Flow rate calculations yield 220 cans, each with a volume of 0.355 l, leading to 78.1 l/min or 1.3 l/s or 0.0013 m³/s.

At Point 2:
A2 = 8 cm² = 0.0008 m²
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

At Point 1:
A1 = 2 cm² = 0.0002 m²
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 =?
Height = 1.35 m

Using Bernoulli’s principle;
P2 + 1/2 * V2² / density = P1 + 1/2 * V1² / density + density * gravitational acceleration * height
=> 152000 + 0.5 * (1.625)² * 1000 = P1 + 0.5 * (6.5)² * 1000 + (1000 * 9.81 * 1.35)
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31 - 34368.5 = 118951.81 Pa = 118.95 kPa
3 0
10 days ago
Read 2 more answers
When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:
Softa [913]

Answer:

option D.

Explanation:

The correct choice is option D.

For an object in equilibrium, the torque measured at any point will be zero.

An object is deemed to be in equilibrium when the net moment acting on it equals zero.

If the object experiences a net moment not equal to zero, it will rotate and will not remain stable.

3 0
10 days ago
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