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3 years ago

9

Explain the difference between waterproofing and dampproofing. When is one or the other an appropriate choice for protecting a f

oundation from moisture?

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

Explanation:

(A) water proofing is designed to keep out soil moisture and liquid water through a concrete structure while damp proofing is a coating on the exterior of a structure so as to stop the transference of ground moisture through concrete.

(B) Water proofing is more appropriate when the water table of the region is higher and there is higher rainfall, it is also more appropriate when there is a presence of hydro-static pressure.

while

Damp proofing is more preferable when there is a drier climate, there is a lower water table, also when there is a well drained soil and also good gravity drain.

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What is another way for

GaryK [48]

Spread or you can say forward

6 0

3 years ago

A car is moving initially at 30 m/s comes gradually at stop at 900 m . What is acceleration of the car ?

ziro4ka [17]

The acceleration of the car is 0.5 meters per seconds square.

Given the following data:

Initial velocity, U = 30 m/s

Distance, S = 900 meters

Note: The final velocity (V) of the car would be zero (0) m/s when it comes to a stop.

To find the acceleration of the car, we would use the third equation of motion;

Mathematically, the third equation of motion is given by the formula;

$V^2 = U^2 - 2aS\\\\0 = 30^2 - 2a(900)\\\\0 = 900 - 1800a\\\\1800a = 900\\\\a = \frac{900}{1800}$

<em>Acceleration, a </em><em>=</em><em> 0.5 </em> $m/s^2$ <em></em>

Therefore, the acceleration of the car is 0.5 meters per seconds square.

Read more: brainly.com/question/8898885

4 0

2 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

6 0

3 years ago

9. In a closed circuit made of a battery, a wire, a lightbulb, and a switch which part's main responsibility is allowing electri

Zielflug [23.3K]

Answer:

d

Explanation:

6 0

3 years ago

Read 2 more answers

IgorC [24]

Answer:

Speed = 300 m/s

Explanation:

Given the following data;

Frequency = 30 Hz

Wavelength = 10 m

To find the speed of the wave;

Mathematically, the speed of a wave is given by the formula:

$Speed = wavelength * frequency$

Substituting into the formula, we have;

$Speed = 10 * 30$

Speed = 300 m/s

6 0

3 years ago