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3 years ago

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the brakes on a car do 240000j of work in stopping the car. if the car travels a distance of 40m while the brakes re being appli

ed. how large is the average force that the brakes exert the car?

1 answer:

Tpy6a [65]3 years ago

3 0

A joule is one Newton of force applied over a meter.
For every meter, the brakes put 240000N of force (N=Newtons).
For 40m, multiply the Newtons by 40.
240000N*40=9600000N

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A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

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3 years ago

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The supporting force exerted by a fluid on an object immersed in it is called ______.

aniked [119]

Answer: buoyant force

Explanation:

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2 years ago

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A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p

Anettt [7]

Answer:

A u = 0.36c B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

u’ (1- u v / c²) = u -v

u + u ’uv / c² = v - u’

u (1 + u ’v / c²) = v - u’

u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

u = 0.51c / (1 + 0.4042)

u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

u = 1.29c / (1- 0.329)

u = 0.961c

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3 years ago

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

son4ous [18]

Answer:

A. 30.7cm

B. $1.7*10^{-10}C$

C. The electric field is directed away from the point of charge

Explanation:

A.

$because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm$

B.

Considering Gauss's law

$EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C$

C. The electric field directed away from the point of charge when the charge is positive.

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A block of mass 3. 0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium, as shown above. The 3. 0 kg block is

QveST [7]

Answer: 32 cm

Explanation:

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