Polar molecules have ionic bonds.<br> True<br> False

I have a massive rock weighing 3,000 Newtons but I can only accelerate it to 500 m/s2 what is its mass?

jeka57 [31]

Answer:

6 kg

Explanation:

F=ma

F is Force(newtons)

m is mass(kg)

a is acceleration(m/s^2)

Plug in the numbers

3000 = m(500)

divide both sides by 500 to cancel out the 500.

3000/500=6

6 = m

6kg is the mass

6 0

3 years ago

Which phenomenon is not the result of thin film interference? Question 15 options: stained glass windows colorful soap bubbles c

katen-ka-za [31]

Rainbows on an oily street after it's been raining. The colors in floating bubbles. Rainbows in the sink when using dish-soap.

3 0

3 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

PLS HELP, WILL GIVE BRAINLIEST!!!

Yuki888 [10]

Answer:

b bro it's b bro

6 0

3 years ago

A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

nikitadnepr [17]

Answer:

The number A(t) of grams of salt in the tank at time t is $A(t) = 150 - 110 e^{-\frac{t}{50} }$

Explanation:

Knowing

$\frac{dA}{dt} = Rin - Rout$

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = ( $\frac{A(t)}{150} g/L$ ) * (3 L/min) = $\frac{A(t)}{50} g/min$

Substituting this results

$\frac{dA}{dt}$ = 3 - $\frac{A(t)}{50}$ --> $\frac{dA}{dt} + \frac{1}{50} A(t) = 4$

Thus, integration factors is

$e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }$

$e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50}$ A(t) = 4 $e^{\frac{t}{50} }$

$e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }$

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

$A(t) = 150 - 110 e^{-\frac{t}{50} }$

8 0

4 years ago

Polar molecules have ionic bonds. True False