Ask question

Ask question

All categories

3 years ago

11

A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 12 N. The carriage has pote

ntial energy. Calculate it.

1 answer:

Ivanshal [37]3 years ago

4 0

The gravitational potential energy is 252 J

Explanation:

The Gravitational Potential Energy (GPE) of an object is given by

$GPE=Wh$

where

W is the weight of the object

h is the height of the object above the ground

For the carriage and the baby in this problem, we have

W = 21 N is their weight

h = 21 m is their height above the ground

Substituting,

$GPE=(12)(21)=252 J$

Learn more about gravitational potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

You might be interested in

An apple from the top branch of the tree and an apple from the bottom branch of a tree fall at the same time. Which apple will h

Vedmedyk [2.9K]

B. The apple from the bottom will hit the ground earlier. This is because an increase in height causes an increase in the time that the object will fall, and therefore will affect the final velocity of the falling object. Moreover, the reduction in velocity due to friction from the air should also be considered.

3 0

3 years ago

Explain what it means when scientists say that the offspring of sexual reproduction are genetically diverse.

Likurg_2 [28]

Answer: During sexual reproduction, the genetic material of two individuals is combined to produce genetically-diverse offspring that differ from their parents. The genetic diversity of sexually-produced offspring is thought to give species a better chance of surviving in an unpredictable or changing environment

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

3 years ago

1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?

harina [27]

Answer:

See step by step sexplanation

Explanation:

1.-Sabemos que la relación:

P₁ * V₁ = P₂ * V₂

Para una temperatura constante debe mantenerse entonces si el globo se comprime hasta llevarlo a 1/3 de su valor inicial, entonces necesariamente para cumplir con la relación mencionada, la presión aumenta tres veces su valor original

2.-La definición de presión es fuerza por unidad de superficie, entonces la fuerza es determinada por la altura de la columna de liquido en el recipiente y no por la cantidad total de liquido, de acuerdo a esto habrá más presión en la base del florero, ya que la columna de agua tiene más altura.

3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque, para el logro del aumento de la presión que es realmente lo que hay que hacer

4 0

3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago