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adoni [48]
3 years ago
12

A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment

um in the northward direction?
Physics
1 answer:
nadya68 [22]3 years ago
5 0
Momentum - mass in motion
P=MV
P=(15,000 kg)(2.5 m/s)
P=37 500 kg x m/s to the north
Hope this helps
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a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
Ber [7]
Fc=mv^2/r so we get 

2000kg*(25m/s)^2/(80m)= 15625N of force 

hope this helps! Thank You!!

4 0
3 years ago
Lab: newton's laws of motion assignment: lab report
poizon [28]

Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

<h3>What are Newton's motion laws?</h3>

Newton's motion laws are a set of scientific statements aimed at explaining the physical property of movement.

These laws explain why objects in movement tend to maintain the same velocity for a short period of time.

In conclusion, Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

Learn more about Newton's motion laws here:

brainly.com/question/10454047

#SPJ1

8 0
2 years ago
A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

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7 0
2 years ago
What os the term for the smallest bit into which chemical substance can be divided and still have the properties of that substan
Ostrovityanka [42]

That's a molecule of the substance.  You can break the molecule down further, into the atoms that make it up, but those don't have the properties of the original  'compound'.

Here's an example:

-- Sodium is a soft, slippery metal, that explodes when water touches it.

-- Chlorine is a poisonous green gas.

When an atom of Sodium and an atom of Chlorine combine, they make one molecule of a substance called "Sodium Chloride".  That's SALT !  It isn't green, it isn't a gas, it isn't poisonous, it isn't soft and slippery, and it doesn't explode when water touches it.

3 0
3 years ago
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