Which process requires the addition of energy to water?

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

3 years ago

Read 2 more answers

What is the magnetic force of a 30m long power line that carries a current of 50A. The magnetic field runs perpendicular to the

Ne4ueva [31]

Answer:

F=BILsin90 when perpendicular sin90 =1 30T x50x30 so you can get 45000N

8 0

3 years ago

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

$r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m$

So, we can re-arrange equation (1) to find the orbital period:

$T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s$

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0

3 years ago

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7) List three (3) automobile safety features currently used to minimize the risk of injury to its passengers. Relate these

r-ruslan [8.4K]

Seatbelt- strapped in egg
air bag- cushion around the egg
brakes- parachute(bag that helps the egg go down slower)

3 0

3 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

4 years ago