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Natalka [10]
3 years ago
9

Hurry!! The first important thing about hiking is that you tell someone where you are going and _________.

Physics
1 answer:
natima [27]3 years ago
3 0

Answer:

when you plan to be back

Explanation:

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(A) State the relation between acceleration and momentum (10 marks).
ohaa [14]

Answer:

momentum is directly proportional to acceleration

4 0
3 years ago
What scientist shot alpha particles into gold
laila [671]
The Geiger–Marsden experiment(s) (also called the Rutherford gold foil experiment) were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where its positive charge and most of its mass are concentrated
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3 years ago
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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
Ivanshal [37]

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

4 0
3 years ago
How much work is needed to push a 117- kg packing crate a distance of 2.75 m up an inclined plane that makes an angle of 28 o wi
Nataly_w [17]

Answer:

W_{tot,min} = 1481.372\,J

Explanation:

The minimum total work is the work needed to counteract the work associated with the weight:

W_{tot, min} = W_{F}

W_{tot,min} = m\cdot g\cdot \sin \theta \cdot \Delta s

W_{tot,min} = (117\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (\sin 28^{\textdegree}) \cdot (2.75\,m)

W_{tot,min} = 1481.372\,J

6 0
3 years ago
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How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water fro
hram777 [196]

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T  +m_w L_v

Here,

m_{Al}= Mass of Aluminum

C_p= Specific Heat of Aluminum

C_w= Specific Heat of Water

m_w = Mass of water

L_v =Latent of Vaporization

Replacing,

\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)

Converting,

\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})

\Delta Q = 440.409kCal

Therefore is required 440.409kCal

8 0
3 years ago
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