Pls help,will give brainliest

Can someone plz tell me how to solve this step by step

avanturin [10]

Answer:

Explanation:

T=2π√l/g

to solve this for l we have to remove the square root in order to finish square root we have to take square on both side so the equation become

T²=4π²l/g

multiplying g on both side

T²g=4π²l

now dividing 4π² on both side

T²g/4π²=l

so l=T²g/4π²

7 0

3 years ago

The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2000 kg. determine the fluid g

leonid [27]

Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2

Thank you for posting your question here at brainly. I hope the answer helps. </span>

4 0

3 years ago

Read 2 more answers

archer shoots his arrow towards a target at a distance of 90 m, and hits ‘bullseye’ . Calculate the acceleration and time taken

mario62 [17]

Answer:

a =45 m/s2

t = 2 seconds

Explanation:

Hi, to answer this question we have to apply the next formula:

v^2 = u^2 +2 a d

Where:

v = final velocity = 90 m/s

u = initial velocity = 0 m/s (shots from rest)

a = acceleration (m/s2)

d = distance = 90m

90^2 = 0^2 + 2a(90)

Solving for a:

8,100= 180 a

8,100/180 = a

a = 45 m/s^2

For time:

v = u + at

90 = 0 + 45t

90/45=t

t =2 seconds

6 0

3 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.