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Ganezh [65]
2 years ago
5

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

Physics
1 answer:
Vitek1552 [10]2 years ago
5 0

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

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A box weighs 100N and its base area of 2 m2. What pressure does it exert on the ground?
kotegsom [21]

Answer:

P = F/S = 100/2 =50 (N/m2)

5 0
2 years ago
What magnet maintains its magnetic proaperties even in the absence of an external magnetic field? a) Ferromagnet b Paramagnet c)
insens350 [35]

Answer:

(a) Ferromagnet

Explanation:

Ferromagnetism is defined as the property by which certain magnets form the permanent magnets.

It is tone of the strong magnetism and it is common phenomenon of magnet in the everyday life of magnetism.

Permanent magnets are made up of ferromagnetic material, in this if the magnetic field is applied then this material is magnetized but do not losses its magnetic property after removal of external magnetic field.

5 0
3 years ago
A 1.0 kg rock is thrown straight upward with an initial speed of 8.0 m/s. What is its speed
Ronch [10]

Answer:5.7m/s

Explanation:

Mass=1kg

Initial velocity=u=8m/s

height=h=1.6m

Final velocity =v

Acceleration due to gravity=g=9.8m/s^2

v^2=u^2-2xgxh

v^2=8^2-2x9.8x1.6

v^2=8x8-2x9.8x1.6

v^2=64-31.36

v^2=32.64

Take the square root of both sides

√(v^2)=√(32.64)

v=5.7

Speed at the height of 1.6m is 5.7m/s

8 0
3 years ago
Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t
ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T_H in the cycle is twice the minimum absolute temperature T_L in the cycle

T_H  = 0.5T_L

now, we find the efficiency of the Carnot cycle engine

η_{th = 1 - T_L/T_H

η_{th = 1 - T_L/0.5T_L

η_{th = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η_{th = 1 - W_{net/Q_H

where W_{net is net work done, Q_H is is the heat supplied

we substitute

0.5 = 60 / Q_H

Q_H = 60 / 0.5

Q_H = 120 kJ

Now, we apply the first law of thermodynamics to the system

W_{net = Q_H - Q_L

60 = 120 -  Q_L

Q_L = 60 kJ

now, the amount of heat rejection per kg of steam is;

q_L = Q_L/m

we substitute

q_L = 60/0.025

q_L = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q_L = h_{fg = 2400 kJ/kg

now, at  h_{fg  = 2400 kJ/kg from saturated water tables;

T_L = 40 + ( 45 - 40 ) ( \frac{2400-2406.0}{2394.0-2406.0}\\} )

T_L = 40 + (5) × (0.5)

T_L = 40 + 2.5

T_L = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C  

4 0
3 years ago
Jill leaves home and rides a distance of 70 km. It took her 2.5 hours. What is her speed?
Paraphin [41]
The answer is ...
28 km per hour
5 0
3 years ago
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