Ask question

Ask question

All categories

liberstina [14]

2 years ago

5

Aunt Jane weights 45 Newtons. What is her mass?

1 answer:

RUDIKE [14]2 years ago

8 0

Answer:

10.116 Pounds/45 newtons = 10.1164024 pounds/force

Explanation:

Divide the newtons by the rate of acceleration, which will give you the mass of the object. The mass will be in kilograms, because a single newton represents the amount of force needed to move one kilogram one meter. For our example, we will divide 10 N by 2 m/s/s, which give us a mass of 5 kg

You might be interested in

Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr

aivan3 [116]

Answer:

The anser is b

Explanation:

4 0

2 years ago

A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the

Oksanka [162]

The average power produced by the soccer player is 710 Watts.

Given the data in the question;

Mass of the soccer player; $m = 67kg$

Energy used by the soccer player; $E = 5100KJ = 5100000J$

Time; $t = 2.0h = 7200s$

Power; $P =\ ?$

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

$Power = \frac{Energy\ converted }{time}$

We substitute our given values into the equation

$Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W$

Therefore, the average power produced by the soccer player is 710 Watts.

Learn more: brainly.com/question/20953664

8 0

2 years ago

0<br> What is 33 C in absolute temperature? hs

AysviL [449]

Answer:

33 Celsius is 306.15 in absolute temperature

7 0

2 years ago

Control rods in a nuclear reactor are used to:

kiruha [24]

B.absorb neutrons to prevent chain reactions which become uncontrollable

7 0

2 years ago

Read 2 more answers

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the

lilavasa [31]

Answer:

$R=803k\Omega$

Explanation:

We have the following information,

$V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s$

We apply the equation for capacitor charging the voltage across it,

$V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}$

Replacing values,

$R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}$

$R=803k\Omega$

3 0

2 years ago