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2 years ago

List three physical properties of a soft drink

Physics

2 answers:

Eva8 [605]2 years ago

8 0

density, amount of gas dissolved, refractive index

topjm [15]2 years ago

4 0

Examples of physical properties include: color, shape, size, density, melting point, and boiling point.

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A scuba diver and her gear displace a volume of 67.0 l and have a total mass of 64.0 kg. (a) what is the buoyant force on the di

slavikrds [6]

Buoyant force is the force that is a result from the pressure exerted by a fluid on the object. We calculate this value by using the Archimedes principle where it says that the upward buoyant force that is being exerted to a body that is immersed in the fluid is equal to the fluid's weight that the object has displaced. Buoyant force always acts opposing the direction of weight. We calculate as follows:

Fb = W
Fb = mass (acceleration due to gravity)
Fb = 64.0 kg ( 9.81 m/s^2)
Fb = 627.84 kg m/s^2

Therefore, the buoyant force that is exerted on the diver in the sea water would be 627.84 N

4 0

3 years ago

If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total

qaws [65]

Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

6 0

3 years ago

A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten

Stella [2.4K]

She can climb 0.92 m without losing weight.

<u>Explanation</u>:

Gravitational potential energy is the energy consisting of the product of mass, gravity and height.

1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J, m = 65.0 kg = 65000 g, Efficiency = 20 %

GPE = mgh

where m represents the mass

g represents the gravity,

h represents the height.

585760 = 65000 $\times$ 9.8 $\times$ h

h = 0.92 m.

7 0

3 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

How is the surface area of an average person is 2m^2 in the chapter pressure

11Alexandr11 [23.1K]

Answer:

We have to show surface area $=2m^2$ ,with few conditions that is by considering Force $=200000\ N$ and Pressure $=100000\ Pa$ to be respectively.