What is the momentum of a car with a mass of 9kg and a velocity of 2m/s?

2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.

WINSTONCH [101]

Yeah un jsjsjsjjsos isnsisoowam

3 0

3 years ago

Convection is usually described as a ___________; when something is heated, it rises, cools, and then falls in a circular patter

creativ13 [48]

Answer:

Current

Explanation:

Convection is the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat. Simply put, Convection is the circular motion that happens when warmer air or liquid — which has faster moving molecules, making it less dense — rises, while the cooler air or liquid drops down.

An everyday example of convection is boiling water ; The heat passes from the burner into the pot, heating the water at the bottom. The water at the bottom rises and is replaced by the water at the top of the pot.

This rise of less dense water at a higher temperature and fall of denser water at a lower temperature sets up a convention current circularly until the water boils. This is a typical example of the day to day application of convection currents.

6 0

3 years ago

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F

muminat

Answer:

$v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

Explanation:

Given that the force of the particle is,

$F_{x}=F_{0}e^{-kt}$

Now it can be further written as

$m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C$

Now the initial conditions are v=1 at t=0.

So,

$1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}$

Now the velocity will become.

$v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

And,

$\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\$

And, another initial condition is x=0 at t=0

$0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}$

Now,

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

5 0

3 years ago

Two blocks A and B have a weight of 11 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of

Andreas93 [3]

Answer:

the block that starts moving first is block A , fr = 1.625 N , fr = 1.5 N

Explanation:

For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane

X axis

fr- Wₓ = 0

fr = Wₓ

Axis y

N- $W_{y}$ = 0

N = W_{y}

Let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

Cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

Wₓ = 11 sin θ

W_{y} = 11 cos θ

The equation for friction force is

fr = μ N

We substitute

μ (W cos θ) = W sin θ

μ = tan θ

We can see that the system began to move the angle.

θ = tan⁻¹ μ

So the angles are

Block A θ = tan⁻¹ 0.15

θ = 8.5º

Block B θ = tan⁻¹ 0.26

θ = 14.6º

So the block that starts moving first is block A

The friction force is

Block A

fr = Wx = W sin θ

fr = 11 sin 8.5

fr = 1.625 N

Block B

fr = 6 sin 14.6

fr = 1.5 N

5 0

3 years ago

an force of 5N is applied to a 10 kg object over a time of 3s. Determine the impulse applied to the object

ddd [48]

Answer:

15Ns

Explanation:

Given parameters:

Mass of the object = 10kg

Force impact = 5N

Time of action = 3s

Unknown:

Impulse applied on the object = ?

Solution:

Impulse is the force that acts on object within a period of time.

Mathematically;

Impulse = Force x time

Now insert the parameters and solve;

Impulse = 5 x 3 = 15Ns

7 0

3 years ago