Ask question

Ask question

All categories

2 years ago

13

I will give brainliest to whoever is first

1 answer:

Roman55 [17]2 years ago

4 0

Answer:

first

Explanation:

also wat u need help with

You might be interested in

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago

If a current-carrying wire is in a magnetic field, in what direction will a force be exerted on the wire?.

astraxan [27]

Answer:

The magnetic force on a current-carrying wire is perpendicular to both the wire and the magnetic field with direction given by the right hand rule.

4 0

1 year ago

A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi

sleet_krkn [62]

Answer:

Equilibrium temperature will be $T=52.2684^{\circ}C$

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So $2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)$

$0.334T-3.67=1688-32.031T$

$T=52.2684^{\circ}C$

5 0

2 years ago

Lipids are ___________ molecules that are insoluble in water. What does this mean?

emmainna [20.7K]

Do you have any options?

8 0

2 years ago

Read 2 more answers

A 120 V fish-tank heater is rated at 130W. Calculate (a) the current through the heater when it is operating, and (b) its resist

7nadin3 [17]

Explanation:

The power P dissipated by a heater is defined as

$P = VI$

where V is the voltage and I is the current.

a) The current running through a 130-W heater is

$I = \dfrac{P}{V} = \dfrac{130\:\text{W}}{120\:\text{V}} = 1.08\:\text{A}$

b) The resistance <em>R</em><em> </em>of the heater is

$P = VI = (IR)I = I^2R$

where $V= IR$ is our familiar Ohm's Law.

$\Rightarrow R = \dfrac{P}{I^2} = \dfrac{130\:\text{W}}{(1.08\:\text{A})^2}$

$R = 110.8\:Ω$

8 0

2 years ago