#2

A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and t

vladimir2022 [97]

This question is incomplete, the complete question is;

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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

8 0

3 years ago

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se

just olya [345]

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

5 0

3 years ago

Why will interference occur

den301095 [7]

Two waves interfere when they run into each other.

The barrier reflects waves that run straight into it. It acts as a wave source and sends wave pulses back up the page towards the incoming waves.

Imagine a loose string tied to a wall. Someone sends two consecutive pulses along the string towards the wall. The first pulse gets reflected right away. It will travel backward towards the person holding the string. Along its way, it will run into the second pulse. The two pulses will interfere. The wall will make the reflected pulse out of phase with the second one. They will end up creating a destructive interference.

So is the case with the water waves running into the barrier. The barrier will send incoming waves back toward where they came from. Reflected waves interfere with incoming ones.

7 0

3 years ago

Now put that book 5 feet in the air. What kind of energy does the book have? Explain.

erastova [34]

Answer:

Gravitational potential energy

Explanation:

The book is put 5 feet in the air, which means 5 feet above the ground. An object which is located to a certain height above the ground possesses a form of energy called gravitational potential energy, which is the energy due to the fact that the object has "potential" to transform this energy into other forms of energy (e.g. kinetic energy, if the book is released and it starts moving).

The value of the gravitational potential energy of the object is given by the formula:

$U=mgh$

where

m is the mass of the object

g is the gravitational acceleration (9.8 m/s^2)

h is the height of the book above the ground (in this case, 5 feet)

So, we see that the gravitational potential energy is proportional to both the mass and the height of the object.

8 0

4 years ago

A bullet is fired from a rifle that is held 3.00 m above the ground in a horizontal position. The initial speed of the bullet is

zloy xaker [14]

Answer: a) 0.78 s; b) 899.83 m; c) Δx=1150*Δt

Explanation: In order to solve this problem we have to use the kinematic equations for the independent two axis (x-y). The following expressions are:

y=yo+voy-(g*t^2)/2 g is 9.8 the aceletarion of the gravity on the y axis

vfy=voy-g*t

x=xo+vox+(a*t^2)/2 a=o there is not acceleration in the x axis.

vfx=vox+a*t

voy and vox ara the initial velocities for each axis.

yo and xo are the initial positions of teh bullet.

From these equations we can obtain the time

y=0=3-(g*t^2)/2

t^2=3*2/g

t= $\sqrt{6/g}$

t= 0.78 s

To calculate xf= voy*t= 1150 * 0.78= 899.83 m

Finally considering with the last equation we ontain the error in Δx

in the form:

Δx=1150*Δt since is linear the dependance with time.

7 0

3 years ago