A 2018 Tesla Model S has a horsepower of 518. How many watts of power does this car model have?

A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises

atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

3 years ago

Read 2 more answers

In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.

alisha [4.7K]

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case $\Delta U=250kJ$ , $\Delta Q$ is the heat tranferred, and $W$ is the work, $W=-480kJ$ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

$\Delta Q=\Delta U +W$

And replacing the known values:

$\Delta Q=250kJ +(-480kJ)$

$\Delta Q=250kJ -480kJ$

$\Delta Q=-230kJ$

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0

3 years ago

On my bike I was able to travel 40 miles in one hour from this information we can determine the bikes

valentinak56 [21]

You have a distance and a time so you can work out the bikes speed which would be distance/time so 40mph.

4 0

3 years ago

Si pudieras viajar a la Luna:

Nostrana [21]

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

4 0

3 years ago