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Andre45 [30]
3 years ago
14

Peak hydraulic systems demands may be met by the use of what ?

Physics
2 answers:
nikklg [1K]3 years ago
8 0
By the use of a Accumulators
Ne4ueva [31]3 years ago
7 0

Answer:

<u><em>The answer is</em></u>: <u>Variable displacement pump.</u>

<u />

Explanation:

The pump <em>increases the hydraulic pressure to the nominal value required by the system. </em>

<u>Variable displacement pump</u>: <em>A variable displacement pump has a fluid outlet that varies to meet the system's pressure demands. The pump output is automatically changed to a compensating pump inside the pump. </em>

Variable displacement pumps are independent of the number of revolutions per minute of the pump as the amount of liquid displaced depends on the needs of the system.

<u><em>The answer is</em></u>: <u>Variable displacement pump.</u>

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A car travels 60.0 miles per hour. What is its velocity in m/s? (1 km = 0.621 mi). . A] 0.01 m/s. .B] 0.03 m/s. .C] 9.66 m/s. .
muminat
By converting miles in to meter 
60 miles=96560 meter
1 hour = 3600 seconds
velocity= 96560m/3600s
Velocity= <span>26.8 m/s</span>
7 0
4 years ago
Read 2 more answers
What is the UL measurement?
liubo4ka [24]
The UL measurement is a symbol used by the International System of Units to represent a microliter. A microliter is a metric unit of measurement for liquid volume and is equal to 1/1,000,000 (one-millionth) of a liter.
4 0
4 years ago
when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer
vazorg [7]
Acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2
7 0
3 years ago
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If the space station is 190 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim?
ozzi

Answer:

The angular velocity produced is 0.321 rad/s.

Explanation:

Given :

Diameter of space station , D = 190 m.

Therefore, radius , R=\dfrac{D}{2}=\dfrac{190}{2}=95\ m.

Also, acceleration , a=9.8\ m/s^2.

We know, angular velocity , \omega=\sqrt \dfrac{a}{R}.

Putting value of g and R in above equation.

We get ,

\omega=\sqrt \dfrac{9.8\ m/s^2}{95\ m}

\omega=0.321\ rad/s.

Hence, this is the required solution.

6 0
3 years ago
Charged particles q1=− 4.80 nC and q2=+ 4.80 nC are separated by distance 3.00 mm , forming an electric dipole. The charges are
Dafna1 [17]

Answer:

Electric field, E = 936.19 N/C

Explanation:

It is given that,

Charge 1, q_1=-4.8\ nC=-4.8\times 10^{-9}\ C

Charge 2, q_2=+4.8\ nC=+4.8\times 10^{-9}\ C

Distance between them, d = 3 mm = 0.003 m

Torque, \tau=8\times 10^{-9}\ N-m

Angle between electric field and line connecting the charge, \theta=36.4^{\circ}

We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :

\tau=pE\ sin\theta

p is the dipole moment, p=qd

\tau=qdE\ sin\theta

E=\dfrac{\tau}{qd\ sin\theta}

E=\dfrac{8\times 10^{-9}}{4.8\times 10^{-9}\times 0.003\ sin(36.4)}

E = 936.19 N/C

So, the magnitude of electric field on the dipole is 936.19 N/C. Hence, this is the required solution.

5 0
3 years ago
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