At what point is the car the fastest

A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

Fed [463]

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

6 0

3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclinedat a 35* angle. The block’s initial speed is 10m/s. (Giventhat

IgorC [24]

a) 4.0 m

We can solve this part by writing the equations of motion along the two directions: perpendicular to the slope and parallel to the slope.

Perpendicular to the slope:

$N-mgcos \theta =0$ (1)

where N is the normal reaction, m = 2.0 kg is the mass of the block, g = 9.8 m/s^2 is the acceleration of gravity, $\theta=35^{\circ}$ is the angle.

Parallel to the slope:

$-\mu_k N -mgsin \theta = ma$ (2)

where $\mu_k=0.20$ is the coefficient of friction, and a the acceleration, and where we have chosen up the slope as positive direction, so both forces are negative.

From (1) we get

$N=mg cos \theta$

And substituting into (2), we can find the acceleration:

$-\mu_k mg cos \theta -mgsin \theta = ma\\a=-\mu_k g cos\theta - g sin \theta = -(0.20)(9.8)(cos 35^{\circ})-(9.8)(sin 35^{\circ})=-7.2 m/s^2$

where the negative sign means the direction is down the slope.

Now we can find the distance travelled along the slope by using the SUVAT equation

$v^2-u^2=2ad$

where

v = 0 is the velocity when the block comes to rest

u = 10 m/s is the initial velocity

d is the distance travelled along the slope

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(10)^2}{2(-7.2)}=6.9 m$

And so, the vertical heigth gained by the block is

$h=d sin \theta = (6.9)(sin 35^{\circ})=4.0 m$

b) 7.4 m/s

The equation of motion along the direction parallel to the slope in this case is

$-\mu_k N +mgsin \theta = ma$

where this time we have taken down the slope as positive direction, so the component of the weight is positive while the frictional force is negative since the block slides downward (while friction acts upward). Solving for a, we find the new acceleration:

$a=-\mu_k g cos \theta + g sin \theta = -(0.20)(9.8)(cos 35^{\circ})+(9.8)(sin 35^{\circ})=4.0 m/s^2$

Now we can use again the SUVAT equation

$v^2-u^2=2ad$

where

v is the final velocity

u = 0 is the initial velocity

d = 6.9 m is the distance travelled along the slope

a = 4.0 m/s^2 is the acceleration

Solving for v,

$v=\sqrt{u^2+2ad}=\sqrt{0+2(4.0)(6.9)}=7.4 m/s$

7 0

3 years ago

A yo-yo is swung around the head of a person in a horizontal circle. What provides the

zhuklara [117]

Answer:

There is a net force towards the center of the circle that causes the yo-yo to seek the center.

Circular Motion Qualities:

-Constant speed

-Accelerating

-Velocity is changing constantly.

Centripetal acceleration perpendicular to the motion causes the object to change direction

a_c=v^2/r

Tangential acceleration in the direction of the motion causes and increase or decrease in speed.

8 0

3 years ago

An object is moving forward and then gets knocked sideways, what is this force called?

Luda [366]

Answer:

It is called force.

Explanation:

A push or pull exerted on an object. Measured in Newtons. Friction & gravity are forces. There are balanced and unbalanced forces.

5 0

3 years ago

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance

PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let $v_h$ be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let $v_v$ be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

$t = 95 / v_h$ or $v_h = 95/t$

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

$s = v_vt + gt^2/2= 0$