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IRISSAK [1]
3 years ago
13

State the relation between acceleration and momentum​

Physics
1 answer:
polet [3.4K]3 years ago
8 0

Answer:

Acceleration is the rate of change in velocity. Momentum is the mass times the velocity. So if you multiply the mass times the acceleration, you get the of change of momentum.

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An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov
jeka94

Answer:

Final speed of boat + man is 1.66 m/s

Explanation:

As we know that there is no friction on the system or there is no external force on this system

So here we can use momentum conservation here

mv = (m + M)v_f

so we have

m = 85 kg

M = 135 kg

v = 4.30 m/s

now we have

85 \times 4.30 = (85 + 135) v

v = 1.66 m/s

4 0
3 years ago
Question 2 (10 points)
kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

F=ma

where

m = 1390 kg is the mass of the truck

a=-3.6 m/s^2 is the acceleration

And substituting,

F=(1390)(-3.6)=-5004 N

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
Which of the following terms is given to a pair of stars that appear to change position in the sky, indicating that they are orb
Ad libitum [116K]

Answer:

option A

Explanation:

The correct answer is option A

The binary star system is the system in which two stars are continuously orbiting each other,

In the eclipsing binary system, two stars revolve about there center of mass and in this system one one-star eclipse another star.

Spectroscopic binary stars are found from the observation of radial velocity and the brighter member of such binary can be seen to have continuously changed the wavelength and periodic velocity.

When the pairs of stars appear to change position in the sky then it is known as visual binary.

5 0
3 years ago
An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act
seropon [69]
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector  \hat{i}.
The positive y-axis = the northern direction, with unit vector \hat{j}.

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})

The plane's actual velocity is the vector sum of the two velocities. It is
\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.
8 0
3 years ago
A two-stage rocket is traveling at 1200 mis with respect to the earth when the first stage runs out of fuel. Explosive bolts rel
Aloiza [94]

Answer:

The speed of the second stage after separation is 4905 m/s

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the system first stage - second stage must remain constant before and after the separation. The momentum of the system is calculated by adding the momentums of each object:

initial momentum = final momentum

m₁₊₂ · v = m1 · v1 + m2 · v2

Where:

m₁₊₂ = mass of the two stage rocket

v = velocity of the rocket

m1 = mass of stage 1

v1 =  velocity of stage 1

m2 = mass of stage 2

v2 = velocity of stage 2

We have the following data:

m1 = 3 · m2

m₁₊₂ = m1 + m2 = 3 · m2 + m2 = 4 · m2

v = 1200 m/s

v1 = -35 m/s  (let´s consider the backward direction as negative)

v2 = ?

Then, replacing these data in the equation of momentum of the system:

m₁₊₂ · v = m1 · v1 + m2 · v2

4 m2 · 1200 m/s = 3 m2 · (-35 m/s) + m2 · v2

Let´s solve the equation for v2:

divide both sides of the equation by m2:

4 · 1200 m/s = 3 · (-35 m/s) + v2

4800 m/s = -105 m/s + v2

v2 = 4800 m/s + 105 m/s = 4905 m/s

The speed of the second stage after separation is 4905 m/s

6 0
3 years ago
Read 2 more answers
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