What is the voltage drop across the 10.0 2 resistor?

2. A hydraulic lift is used to lift heavy machine pushing down on a 5 square meters piston with a force of 1000 N. What force ne

IRINA_888 [86]

Answer:

The force needs to be applied on the 1 square meter piston to lift the machine is 200 N.

Explanation:

Given that,

Force, $F_1=1000\ N$

Area, $A_1=5\ m^2$

We need to find the force needs to be applied on the 1 square meter piston to lift the machine. We know that the pressure at every point is same due to Pascal's law such that :

$P_1=P_2\\\\\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\\\dfrac{1000}{5}=\dfrac{F_2}{1}\\\\F_2=200\ N$

So, the force needs to be applied on the 1 square meter piston to lift the machine is 200 N. Hence, this is the required solution.

3 0

4 years ago

Read 2 more answers

Earthquakes are more prevalent along fault lines<br> A. true<br> B. false

Umnica [9.8K]

A. True, for example there are more earthquakes in California because of the St. Andreas fault

8 0

4 years ago

If 2 objects have the same momentum which statement is true

Vika [28.1K]

C is the correct option

As momentum = (mass) (velocity)

P=mv

So the object having greater mass has less velocity and the object having smaller mass will move faster. But product of mass and velocity is same to each other.

5 0

3 years ago

Suppose that the period of a particular ideal mass-spring system is 5 s . What would be the period of the system if the mass wer

yan [13]

Answer: T2 = 7.07s

Explanation: The period of a loaded spring of spring constant k and mass m is given by

T= 2π √m/k

With 2π constant and k, it can be seen with little algebra that

T² is proportional to mass m

Hence (T1)²/m1 = (T2) ²/m2

Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.

By substituting, we have that

5²/m = (T2) ²/2m

25 / m = (T2) ²/2m

25 × 2m = (T2) ² × m

25 × 2 = (T2) ²

50 = (T2) ²

T2 = √50

T2 = 7.07s

5 0

4 years ago

If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the

koban [17]

Answer:

$\displaystyle \frac{480}{7}\approx 68.6\; \rm mph$ .

Explanation:

The average speed of an object is equal to total distance over total time.

Distance traveled: $\rm 50 \; mi$ .

How much time is taken? This trip is divided into two halves, each of distance $\displaystyle \frac{50}{2} = 25\;\rm mi$ .

Time spent on the first half of the trip:

$\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours$ .

Similarly, time spent on the second half of the trip:

$\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours$ .

In total:

$\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours$ .

Average speed:

$\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}$ .

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to $\rm 60\; mph$ rather than $\rm 80\; mph$ .

3 0

3 years ago