Ask question

Ask question

All categories

2 years ago

9

Which is not a characteristic of an ideal fluid?

1 answer:

Lena [83]2 years ago

7 0

The answer would be B

You might be interested in

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg

marissa [1.9K]

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

3 0

2 years ago

A cart is pushed to the right with a force of 15 N while being pulled to the left with a force of 20 N. The net force on the car

9966 [12]

The net force of the cart when it is pushed to the right with a force of 15N.

<u>Explanation:</u>

To find the force of net, which is calculated by the formula.

The Net Force= Addition of the force applied on the respective direction.

The Net Force here is given by

The Net Force = 15-20 (A force towards the right and a force towards left, two opposite so subtraction).

Hence

Thus the Net Force = -5(The force towards left, so it gets a negative value).

5 0

2 years ago

A cylindrical specimen of brass that has a diameter listed above, a tensile modulus of 110 GPa, and a Poisson's ratio of 0.35 is

Irina-Kira [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The strain experienced by the specimen is 0.00116 which is option A

Explanation:

The explanation is shown on the second uploaded image

8 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago

The Olympic record for running the 200m dash is 19.3 seconds. What is the average speed for this record?

Aloiza [94]

I think it’s c. if not i’m sorry!!

3 0

2 years ago