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2 years ago

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a driver shifts into neutral when her 1200 kg is moving at 80 km/h and finds the speed has dropped to 65 km/h 10 s later . what

was the average drag force acting on the car? ...?

1 answer:

Liono4ka [1.6K]2 years ago

6 0

Drag<span> is a </span>force<span> acting opposite to the relative motion of any object moving. It can be calculated as follows:

</span><span><span>vi </span>= (<span><span>80km/</span><span>hr) (1</span></span><span><span>hr / </span><span>3600sec) (</span></span><span><span>1000m/</span><span>1km) </span></span>= <span><span>22.2 m/</span>sec

</span></span> <span><span>vf</span>= (<span><span>65km/</span><span>1hr) (</span></span><span><span>1hr/</span><span>3600sec) (</span></span><span><span>1000m/</span><span>1km) </span></span>= <span><span>18.06 m / </span>sec</span></span> <span>
a=<span><span><span>vf</span>−<span>vi / </span></span><span><span>tf</span>−<span>ti = </span></span></span><span><span>18.06m/s−22.2m/s / </span><span>10sec−0sec</span></span>= <span><span>−0.414m</span><span>sec2

</span></span></span><span>F = ma = 1200kg×<span><span>−0.414m</span><span>s2 </span></span>= −496.8N</span>

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Answer: The things we know are:

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Sam, the possible father, is a yellow labrador.

Discarding things such as a scientific indagation, the breeder could look for traits in the pups.

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A good starting point is looking at the color, the posture, and things like that.

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2 years ago

Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?

morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays, parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

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2 years ago

How many slices of bread did each climber have to eat to compensate for the increase of the gravitational potential energy of th

N76 [4]

Answer:

So No of slices to be consumed by each person = n = 65

Explanation:

Energy released by one slice = E1

$E1=10^6\ J$

h = 8850 m ; m = 79 kg ,η= 10.5%

We know that potential energy given as

u = m g h

u = 79 x 9.81 x 8850

$u=6.8\times 10^6\ J$

we know from the defination of efficiency that, η= E(out)/E(in)

Now amount of PE has to be compensated, In our case, E(out) =u

$0.105=\dfrac{E(out)}{E(in)}$

$0.105=\dfrac{6.8\times 10^6}{E(in)}$

$E(in)=64.76\times 10^6\ J$

Let n be the number of bread slices to be consumed.

n = E(in)/E1

$n=\dfrac{64.76\times 10^6}{10^6}$

n=64.76

So No of slices to be consumed by each person = n = 65

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2 years ago

A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from

Naddik [55]

Complete Question

The complete question iws shown on the first uploaded image

Answer:

a

$y \to z \to x$

b

$x \to z \to y$

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

Now the magnetic force is mathematically represented as

$F = I L$ x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

The motion of the loop is from y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

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5 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

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

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

b)

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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2 years ago