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liq [111]
3 years ago
13

What to forces effects the momentum of an object

Physics
2 answers:
noname [10]3 years ago
7 0
<span>Any force can affect the momentum of an object</span>
faltersainse [42]3 years ago
4 0

Any force can affect the momentum of an object
if it changes the object's speed.

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A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph
Marina86 [1]

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

8 0
3 years ago
A cube slides down the surface of a ramp at a constant velocity. What is the magnitude of the
Stella [2.4K]
A . The weight of the cube
7 0
3 years ago
A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina
elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec        

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car a=1.6m/sec^2 in negative direction so acceleration will be a=-1.6m/sec^2

From first equation of motion we know that

v = u+at

So v=0+(-1.6)\times 5.8=-9.28m/sec

So final velocity will be -9.28 m/sec

8 0
3 years ago
In 2.8 s, a car increases its velocity from 20. M/s to 25 m/s. What is the acceleration of the car
Akimi4 [234]
Acceleration= change in velocity/time; 5/2.8 , so a=1.785714286
6 0
3 years ago
Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob
pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

T^2 = \frac{4\pi^2}{GM}a^3

Where

T= Period

G= Gravitational constant

M = Mass of the sun

a= The semimajor axis of the comet's orbit

The period in years would be given by

T= 1941-550\\T= 1391y(\frac{31536000s}{1y})\\T=4.3866*10^{10}s

PART A) Replacing the values to find a, we have

a^3= \frac{T^2 GM}{4\pi^2}

a^3 = \frac{(4.3866*10^{10})^2(6.67*10^{-11})(1.989*10^{30})}{4\pi^2}

a^3 = 6.46632*10^{39}

a = 1.86303*10^{13}m

Therefore the semimajor axis is 1.86303*10^{13}m

PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by

R = a(1-e)

R = 1.86303*10^{13}(1-0.997)

R= 5.58*10^{10}m

7 0
3 years ago
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