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jenyasd209 [6]
3 years ago
7

If a car has a mass of 1,000 kilograms, and a velocity of 35 m/s. What is its momentum?

Physics
1 answer:
fenix001 [56]3 years ago
7 0

\vec{p}=m\vec{v}\\p = 1000 * 35=35000[\frac{kg*m}{s}]

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A 300-kg bear grasping a vertical tree slides down at constant velocity. the friction force between the tree and the bear is:___
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3000 N of friction exists between the bear and the tree.

The bear goes down at a constant speed, thus there is no acceleration. The forces add up to zero. As a result, upward friction and downward weight result in zero  

i.e.,

F-mg=0

F = mg

F = (300 kg)/10 m/s^{2} (gravitational acceleration)

3000 N frictional force equals F.

How does friction force work?

  • Friction is the force that prevents two solid objects from rolling or sliding over one another. Even though frictional forces, such as the traction required to walk without slipping, may be advantageous, they can also be a significant hindrance to motion.

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Fossils found in the La Brea tar pits indicate a California climate that was A) similar to today's climate. B) similar to the pr
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Answer:

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Explanation:

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Nina is doing a workout where she runs at 50% speed for 3 minutes, 75% speed for 2 minutes, and 100% speed for 30 seconds. This
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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o
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Answers:

(a) 0.0073kg

(b) Volume gold: 3.79(10)^{-7}m^{3}, Volume cupper: 7.6(10)^{-8}m^{3}

(c) 17633.554kg/m^{3}

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass M of the coin, which is an alloy  of gold and copper is:

M=m_{gold}+m_{copper}=7.988g=0.007988kg   (1)

Where  m_{gold} is the mass of gold and m_{copper} is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats K and mass is:

K=24\frac{m_{gold}}{M}   (2)

Finding {m_{gold}:

m_{gold}=\frac{22}{24}M   (3)

m_{gold}=\frac{22}{24}(0.007988kg)   (4)

m_{gold}=0.0073kg   (5)  This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density \rho of an object is given by:

\rho=\frac{mass}{volume}

If we want to find the volume, this expression changes to: volume=\frac{mass}{\rho}

For gold, its volume V_{gold} will be a relation between its mass m_{gold}  (found in (5)) and its density \rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}:

V_{gold}=\frac{m_{gold}}{\rho_{gold}}   (6)

V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}   (7)

V_{gold}=3.79(10)^{-7}m^{3}   (8)  Volume of gold in the coin

For copper, its volume V_{copper} will be a relation between its mass m_{copper}  and its density \rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}:

V_{copper}=\frac{m_{copper}}{\rho_{copper}}   (9)

The mass of copper can be found by isolating m_{copper} from (1):

M=m_{gold}+m_{copper}  

m_{copper}=M-m_{gold}  (10)

Knowing the mass of gold found in (5):

m_{copper}=0.007988kg-0.0073kg=0.000688kg  (11)

Now we can find the volume of copper:

V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}   (12)

V_{copper}=7.6(10)^{-8}m^{3}   (13)  Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density \rho_{coin will be a relation between its total mass M and its total volume V:

\rho_{coin}=\frac{M}{V} (14)

Knowing the total volume of the coin is:

V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3} (15)

\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}} (16)

Finally:

\rho_{coin}=17633.554kg/m^{3}} (17)  This is the total density of the British sovereign coin

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