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Musya8 [376]
3 years ago
14

What is the momentum of a 31.2 kg object traveling at a velocity of 2.1 m/s?

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

Answer:

p = 65.52 kg m/s

Explanation:

p = m × v

p = 31.2 × 2.1

p = 65.52 kg m/s

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Newton's first law of motion.
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Which process is involved in the formation of a galaxy
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A 0.150-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
Hitman42 [59]

Answer:

The magnitude of the change in momentum of the stone is 5.51kg*m/s.

Explanation:

the final kinetic energy = 1/2(0.15)v^2

                1/2(0.15)v^2  = 70%*1/2(0.15)(20)^2

                              v^2 = 21/0.075

                              v^2 = 280

                                 v = 16.73 m.s

if u is the initial speed and v is the final speed, then:

u = 20 m/s and v = - 16.73m/s

change in momentum = m(v-u)

                                     = 0.15(- 16.73-20)

                                    = -5.51 kg*m.s

Therefore, The magnitude of the change in momentum of the stone is 5.51kg*m/s.

3 0
3 years ago
A man walks 15 meters in 15 seconds. what is his speed​
ddd [48]

Answer:

1 meter per second

Explanation:

Every second, he walks 1 meter because 15 ÷ 15 = 1

3 0
3 years ago
To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad
Julli [10]

The magnitude of the forces acting at the top are;

\mathbf{F_{Top, \ x}} = 132.95 N

\mathbf{F_{Top, \ y}} = 0

The magnitude of the forces acting at the bottom are;

\mathbf{F_{Bottom, \ x}} = \mathbf{ F_f} = -132.95 N

\mathbf{F_{Bottom, \ y}} = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, <em>B </em>gives;

\sum M_B = 0

Therefore;

\sum M_{BCW} = \sum M_{BCCW}

Where;

\sum M_{BCW} = The sum of clockwise moments about <em>B</em>

\sum M_{BCCW} = The sum of counterclockwise moments about <em>B</em>

Therefore, we have;

\sum M_{BCW} = 2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

\sum M_{BCCW} = F_R × √(6² - 2²)

Therefore, we get;

2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81  = F_R × √(6² - 2²)

F_R  = (2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, F_R ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, F_R = The magnitude of the frictional force of bottom of the ladder on the floor, F_f but opposite in direction

Therefore;

F_R = -F_f

F_f = - F_R ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = \sum F_y = 0

The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;

\sum F_y = -70.0 × 9.81 - 10 × 9.81 + F_{By}

∴ The upward force acting at the bottom, F_{By} = 784.8 N

Therefore;

The magnitudes of the forces at the ladder top and bottom are;

At the top;

\mathbf{F_{Top, \ x}} = F_R ≈ 132.95 N←

\mathbf{F_{Top, \ y}} = 0 (The surface upon which the ladder rest at the top is frictionless)

At the bottom;

\mathbf{F_{Bottom, \ x}} = F_f ≈ -132.95 N →

\mathbf{F_{Bottom, \ y}} = F_{By} = 784.8 N ↑

Learn more about equilibrium of forces here;

brainly.com/question/16051313

8 0
3 years ago
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