Ask question

Ask question

All categories

3 years ago

8

If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter what’s the

process at point A
A. 9.80 kPa
B. 3.85 kPa
C. 7.84 kPa
D. 5.88 kPa

1 answer:

LenaWriter [7]3 years ago

4 0

The answer is C I believe

You might be interested in

What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i

Katarina [22]

R = 0.407Ω.

The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0

2 years ago

A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed $v_1=8 m/s$

initial angular velocity $\omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s$

Speed after 4.1 s is $v_2=2.2 m/s$

$\omega _2=\frac{2.2}{0.6}=3.66 rad/s$

using $\omega _2=\omega _1+\alpha t$

where $\alpha$ is angular acceleration

$3.66=13.33+\alpha \cdot 4.1$

$\alpha =-2.37 rad/s^2$ i.e. clockwise

(b)angular displacement

$\theta =\omega _1t+\frac{\alpha t^2}{2}$

$\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}$

$\theta =54.66-19.75$

$\theta =34.91 rad$

3 0

3 years ago

Current is the movement of positive charges called electrons

GaryK [48]

Answer:what

Yeah

Explanation:

6 0

2 years ago

Read 2 more answers

A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe

pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

$E_p = mgh$

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

$E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J$

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

$E_p = E_k = kx^2/2 = 13x + 14.3$

where k = 315 is the spring constant and x is the compressed length

$315x^2 = 26x + 28.6$

$315x^2 - 26x - 28.6 = 0$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}$

$x = \frac{26 \pm 191.6}{630}$

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m

6 0

3 years ago

Genes are located within a cell’s BLANK. Each gene controls the production of one or more specific BLANK. Changes in the structu

Thepotemich [5.8K]

Answer:

Genes are located in the CHROMOSOME

Explanation:

chromosome is were genes are located

7 0

2 years ago