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3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Physics

1 answer:

Anika [276]3 years ago

3 0

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a

devlian [24]

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + \frac{1}{2}(-g)t^2$

=> $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^{-1}[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

6 0

3 years ago

Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below

MatroZZZ [7]

The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

Learn more about height on:

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8 0

2 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

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2 years ago

Which of the following best describes a plane?

katrin [286]

B, the surface of a flat table.

4 0

3 years ago

Read 2 more answers

Technician A says when diagnosing an overheating hydraulic system, be sure that the oil cooler is not plugged and the cooler’s f

mel-nik [20]

Answer:

Tech B is correct and Tech A is incorrect.

Explanation:

Here Tech A is wrong because when diagnosing an overheating hydraulic system, it is not necessary to un plugg the oil cooler rather it should be plugged to for proper diagnosis.

Technician B says running the hydraulic system at a lower operating temperature will reduce the possibly of oil oxidation is correct statement because at temperature oil's physical as well as chemical property tend to change.

Hence, Tech B is correct and Tech A is incorrect.

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3 years ago