Witch of the following questoons would you expect to see on an interest inventory

A car that increase its speed from 20 km/h to 100 km/h undergoes -------acceleration,

stepladder [879]

negative acceleration- deceleration

8 0

3 years ago

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

3 years ago

Question 1<br> 2.5 cm=<br> mm

Nookie1986 [14]

There’s 10mm in a cm: 22mm

8 0

3 years ago

Read 2 more answers

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

$T = \frac{2u Sin\theta }{g}$