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3 years ago

2. Find the weight of a person (on Earth) in Newtons if they have a mass= 55.0 kg.

Physics

1 answer:

adelina 88 [10]3 years ago

8 0

Answer:

539 N

Explanation:

Fg = mg

Step 1: Define

Fg = ?

m = 55.0 kg

g = 9.8 m/s²

Step 2: Substitute and Evaluate

Fg = 55.0 kg(9.8 m/s²)

Fg = 539 N

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What happens when a star exhausts its core hydrogen supply?

Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

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3 years ago

Given the equation E = P/N , solve for P

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Answer:

E=P/N

multiply both sides by N

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3 years ago

A force of 250 N is applied to a 1 kg softball when struck with a bat. what is the acceleration

My name is Ann [436]

A=f/m
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a=250m/s^2

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4 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

mrs_skeptik [129]

Answer:

0.25 L

Explanation:

$P_1$ = Initial pressure = 1 atm

$T_1$ = Initial Temperature = 20 °C

$V_1$ = Initial volume = 4.91 L

$P_2$ = Final pressure = 5.2 atm

$T_2$ = Final Temperature = -196 °C

$V_2$ = Final volume

From ideal gas law we have

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L$

The pressure experienced by the balloon is 0.25 L

7 0

3 years ago