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adelina 88 [10]

3 years ago

9

If your breaks fail slam on them as hard as possible

2 answers:

9966 [12]3 years ago

7 0

False hope this helped

ikadub [295]3 years ago

5 0

Yeah this is the false

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John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

7 0

3 years ago

Which change would increase the electric force between two objects

ladessa [460]

Answer:

<em>Reducing the distance between the charges increases the electric force</em>

Explanation:

<u>Electrostatic Force </u>

The force between two point charges can be computed by using the Coulomb's law formula:

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

where k is a constant, q1 and q2 are the charges and r is the distance between them. If the charges have a constant value, if we wanted to increase the force between them, we must get them closer to each other. For example, if the distance reduces by one-half, the new force will be

$\displaystyle F'=\frac{k\ q_1\ q_2}{(r/2)^2}$

$\displaystyle F'=\frac{k\ q_1\ q_2}{r^2/4}$

$\displaystyle F'=4\frac{k\ q_1\ q_2}{r^2}$

Or equivalently

$F'=4F$

Reducing the distance to a half, the force will increase to 4 times

8 0

3 years ago

At a certain instant, the earth, the moon, and a station- ary 1250-kg spacecraft lie at the vertices of an equilateral triangle

Murrr4er [49]

Answer:

a). F = 3.376 N, θ = 59.18°

b). W = 1.3x $10^{9}$ J

Explanation:

We know

Gravitational constant, G = 6.673 x $10^{-11}$ N- $m^{2}$ / $kg^{-2}$

Mass of the earth, M = 5.97 x $10^{24}$ kg

mass of the moon, m = 7.35 x $10^{22}$ kg

Mass of the satellite, $m_{s}$ = 1250 kg

Distance between the objects, r = 3.84 x $10^{5}$ km

= 3.84 x $10^{8}$ m

Now

The force on the satellite due to moon

$F_{m}= \frac{G\times m\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 7.35\times 10^{22}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 0.0415 N ( in the positive x direction )

The force on the space craft due to the earth

$F_{m}= \frac{G\times M\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 5.97\times 10^{24}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 3.377 N ( at 60° to x axis )

Now component of force of earth along x axis

$F_{e_{x}} = F_{e}\times cos 60$

= 3.377 x 0.5

= 1.6885 N

Now component of force of earth along y axis

$F_{e_{y}} = F_{e}\times sin 60$

= 3.377 x 0.86

= 2.90 N

∴ Net force on the space craft due to earth and moon along x axis

$F_{x}$ = $F_{e}$ cos 60+ $F_{m}$

= 1.3885+0.0415

= 1.73 N

Net force on the space craft due to earth and moon along y axis

$F_{x}$ = $F_{e_{y}}$

= 2.90 N

Therefore, total force F = $\sqrt{(F_{x}^{2})+(F_{y}^{2})}$

F = $\sqrt{(1.73^{2})+(2.90^{2})}$

F = 3.376 N

∴ Magnitude of the net gravitational force on the space craft is 3.376 N

Direction of net force on the space craft is given by

$\Theta = \arctan \left (\frac{F_{y}}{F_{x}}\right )$

$\Theta = \arctan \left (\frac{2.90}{1.73}\right )$

$\Theta = 59.18$ °

Therefore this direction is 59.18° from the line joining earth and the space craft.

b).

∴ Gravitational potential energy of the space craft is given by

$E = \frac{G.M.m_{s}}{r}+\frac{G.m.m_{s}}{r}$

$E = \frac{G\times m_{s}\left ( M+m \right )}{r}$

$E = \frac{6.673\times 10^{-11}\times 1250\left ( 5.97\times 10^{24}+7.35\times 10^{22} \right )}{3.84\times 10^{8}}$

E = 1312769385 J

E = 1.3 x $10^{9}$ J

Therefore minimum work done is 1.3x $10^{9}$ J

8 0

3 years ago

The Sun delivers an average power of 15.29 W/m2 to the top of Saturn's atmosphere. Find the magnitudes of vector E max and vecto

Elenna [48]

Answer:

$E=\sqrt{\frac{2u_E}{\epsilon_0}}=1.8*10^6N/C\\B=\sqrt{2\mu_0}=1.58*10^{-3}T$

Explanation:

The energy density of electromagnetic waves can be computed as

$u_{E}=\frac{1}{2}\epsilon_0E^2\\u_B=\frac{B^2}{2\mu_0}$

The power is the energy per unit of time. Hence we can take uE and Ub as 15.29J

By taking apart E and B we have

$E=\sqrt{\frac{2u_E}{\epsilon_0}}=1.8*10^6N/C\\B=\sqrt{2\mu_0}=1.58*10^{-3}T$

hope this helps!!

6 0

3 years ago

Read 2 more answers

Sarah invested 12000 in a unit trust 5 years ago, the value of the unit trust has increased by 7% per annum for the last 3 years

miss Akunina [59]

Answer:

The value of the fund today is $13831.72

Explanation:

Let us calculate the value at the end of each year, step by step. An annual decrease of 3% corresponds to a multiplier 1.0-0.03 = 0.97. An increase of 7% corresponds to 1.07.

Value at the end of

year 1: 12000 * 0.97

year 2: 12000 * 0.97 * 0.97

year 3: 12000 * 0.97 * 0.97 * 1.07

year 4: 12000 * 0.97 * 0.97 * 1.07 * 1.07

year 5: 12000 * 0.97 * 0.97 * 1.07 * 1.07 * 1.07 = 13831.72

The value of the fund today (end of year 5) is $13831.72

8 0

4 years ago

Read 2 more answers