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3 years ago

ANSWER PLEASE FAST I NEED IT

Physics

1 answer:

Deffense [45]3 years ago

6 0

14.0.5 cm 13. wind speed surface area temperature

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Calculate the volume of this regular solid. A cylinder labeled B at the top, 13 centimeters high with a radius of 4 centimeters.

Kryger [21]

Answer:

The volume of the cylinder is $653.71\ \text{cm}^3$

Explanation:

Given that,

The height of the cylinder is 13 cm

Radius of the cylinder is 4 cm

We need to find the volume of the cylinder. The formula that is used to find the volume of a cylinder is given by :

$V=\pi r^2 h\\\\V=\dfrac{22}{7} \times (4)^2 \times 13\\\\V=653.71\ \text{cm}^3$

So, the volume of the cylinder is $653.71\ \text{cm}^3$

4 0

3 years ago

Read 2 more answers

An airplane is flying in air with a density of 1.29 kg/m3. A pressure gauge measures

jeka57 [31]

Answer:

341 m/s

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

The velocity of the air at the nose is 0 m/s, so:

P₁ = P₂ + ½ ρ v₂²

ΔP = ½ ρ v₂²

Plugging in values:

75000 Pa = ½ (1.29 kg/m³) v²

v = 341 m/s

5 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

What is the answer?

scoundrel [369]

"Choice-C" is the answer.

7 0

3 years ago

The gravitational force of attraction between two objects would increase by

OleMash [197]

The gravitational force of attraction between two objects would be increased by "decreasing the distance between two objects"

Hope this helps!

6 0

3 years ago