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svp [43]
1 year ago
8

in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse

Physics
1 answer:
Greeley [361]1 year ago
8 0

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

Learn more about impulse here

brainly.com/question/16980676

#SPJ4

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Sck my p3nis

Explanation:

if you do so, then your mom will have coronavirus.

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3 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

7 0
2 years ago
A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the ma
Sergio [31]

Answer:

mgh= 10 x 8 x 10

= 800

but you can try 10 x 8 x 4^-1 x 10

6 0
2 years ago
A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resistin
otez555 [7]

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

Acceleration,

a=3 \ m/s^2

Forces resisting motion,

F_f=432 \ N

Mass,

m = 241 \ kg

By using Newton's second law, we get

⇒ F-F_f=ma

Or,

⇒         F=ma+F_f

By putting the values, we get

⇒             =(3\times 241)+432

⇒             =723+432

⇒             =1155 \ N

7 0
3 years ago
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
2 years ago
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