How much soil is in a 5 cm to 9 cm deep hole?

Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the

Naddik [55]

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

$\Sigma F = F - f = m\cdot a$ (Eq. 1)

Where:

$F$ - Force from the wind exerted on the sailboat, measured in newtons.

$f$ - Drag force of the water, measured in newtons.

$m$ - Mass of the sailboat, measured in kilograms.

$a$ - Net acceleration of the sailboat, measured in meters per square second.

If we know that $F = 3.90\times 10^{3}\,N$ , $f = 4.35\times 10^{3}\,N$ and $m = 1250\,kg$ , then the net acceleration of the sailboat is:

$a = \frac{F-f}{m}$

$a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}$

$a = -\frac{9}{25}\,\frac{m}{s^{2}}$

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

$v = v_{o}+a\cdot t$ (Eq. 2)

Where:

$v_{o}$ , $v$ - Initial and final velocities of the sailboat, measured in meters per second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 11\,\frac{m}{s}$ , $a = -\frac{9}{25}\,\frac{m}{s^{2}}$ and $t = 15\,s$ , then the final velocity of the sailboat is:

$v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}} \right)\cdot (15\,s)$

$v = 5.6\,\frac{m}{s}$

The velocity of the boat 15 seconds later is 5.6 meters per second.

3 0

3 years ago

An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

Tresset [83]

Answer:231.16 N/C

Explanation:

Given

Electric Flux $=147 N-m^2/C$

Area(A) $=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6= $58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C

4 0

3 years ago

A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

AveGali [126]

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

3 0

2 years ago

A woman gets burned at the beach after a hot, sunny day. What kind of

igomit [66]

Answer:

thermal, hot equals thermal and such

3 0

2 years ago

Read 2 more answers

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

3 years ago