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BARSIC [14]
3 years ago
6

You throw a balloon that floats in the air with a velocity of 2 m / s south . If the wind speed is 5 m / s west , how far south

will the balloon travel after 2 seconds ?
Physics
1 answer:
zvonat [6]3 years ago
4 0

Answer:

The distance traveled by the balloon is 10.77 m

Explanation:

velocity of the ball, v_b = 2 m/s south

velocity of the air, v_a = 5 m/s west

To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.

                                       | 2m/s

                                       |

                                       |

                                      ↓

      5m/s  ←------------------

the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.

R² = 5² + 2²

R² = 29

R = √29

R = 5.385 m/s

The distance traveled by the balloon is calculated as;

d = R x t

where;

t is time of the motion = 2 seconds

d = 5.385 x 2

d = 10.77 m

Therefore, the distance traveled by the balloon is 10.77 m.

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The similarity between thermal energy, heat and temperature?
Vitek1552 [10]
Heat energy and thermal energy are the same because heat energy is thermal energy. Also thermal energy and temperature are the same because temperature is measuring heat in degrees Celsius or degrees Fahrenheit. Hope this helps!
6 0
3 years ago
A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
What is the mass of an object with a density of 18 g/cm3 and a volume of 25 cm3
DiKsa [7]

Answer: 0.72 grams

Explanation: Mass can be extracted from the formula of density. D=M/V where D is density and V is volume. Therefore:

18 g/cm^3 = M(25 cm^3) --> Divide by 18g/cm^3 by 25 cm^3 to isolate mass. --> <u>0.72 =M </u> --> Now, to find out which unit you need to use for mass, just look at the density. You can see it is in g/cm^3, and cm^3 was already used for the volume. Thus, gram units are left, so that will be the unit needed, making the final answer 0.72 grams. Hope this helps :)

5 0
2 years ago
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off th
Black_prince [1.1K]

Answer:

4units

Explanation:

To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;

D = √(x2-x1)²+(y2-y1)²

Given the coordinate points

(3,5) and (7,5)

Substitute

D = √(7-3)²+(5-5)²

D = √(7-3)²+0²

D = √4²

D = √16

D = 4

Hence the total distance the beam will travel along this path is 4units

6 0
3 years ago
A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe
Allushta [10]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

P = \sigma Ae\Delta T^4

Where

A = Surface area of the Object

\sigma = Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

A = 1.36m^2

\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4

\sigma = 5.67*10^{-8} J/(s m^2 K^4)

P = 122J/s

e = 0.7

Replacing at our equation and solving to find the temperature 1 we have,

P = \sigma Ae\Delta T^4

P = \sigma Ae (T_2^4 -T_1^4)

122 = (5.67*10^{-8})(1.36)(0.7)(307^4-T_1^4)

T_1 = 285.272K = 12.122\°C

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

8 0
3 years ago
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