All categories

2 years ago

A 55 kg person falling with a velocity of 0.6

Physics

1 answer:

siniylev [52]2 years ago

3 0

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

You might be interested in

A resistor with an unknown resistance is connected in parallel to a 13 ? resistor. when both resistors are connected in parallel

DENIUS [597]

Hope this helps you.

4 0

2 years ago

My mass is 65 kg and on Earth this equals a weight of 640 N, but on the moon where gravity is 1.7 m/s² my

Helen [10]

Your weight on the moon given the data from the question is 110.5 N

<h3>Definition of mass and weight </h3>

Mass is simply defined as the quantity of matter present in an object. The mass of an object is constant irrespective of the location of the object.

Weight is simply defined as the gravitational pull on an object. The weight of an object varies from place to place due to gravity.

<h3>Relationship between mass and weight </h3>

Mass and weight are related according to the following equation

Weight (W) = mass (m) × Acceleration due to gravity (g)

<h3>How to determine the weight on the moon</h3>

Mass (m) = 65 Kg
Acceleration due to gravity on the moon (g) = 1.7 m/s²
Weight (W) =?

W = mg

W = 65 × 1.7

W = 110.5 N

Learn more about mass and weight:

brainly.com/question/14684564

#SPJ1

4 0

1 year ago

What never change despite what planet you go to

laiz [17]

Your mass will never change despite if you go to Jupiter, Uranus, Mars, Earth, or any planet.

7 0

3 years ago

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass $m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}$

$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

2 years ago

Which organization functions to privide clinics and surgeries i developing nations

Nimfa-mama [501]

Doctors without boarders.

4 0

2 years ago