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3 years ago

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Need help ASAP ????????

1 answer:

zloy xaker [14]3 years ago

8 0

Answer:

Question 1 is true, question 2 is b

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Momentum is a vector quantity which means it has both ___________________ and _____________

Damm [24]

<span>magnitude and direction</span>

6 0

3 years ago

Read 2 more answers

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

a crane lifts a 75 kg mass at a height of 8m. calculate the gravitational potential energy gained by the mass (g= 10N)

Leno4ka [110]

Answer:

6000 J

Explanation:

Potential energy = mass*gravity*height (U = m*g*h)

In this case g = 10 so

U = 75*10*8 = 6000 J (joules) [kg * m^2/s^2]

7 0

3 years ago

A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.

marin [14]

Answer: A

Explanation:

I want my points so yea

4 0

2 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

4 years ago