Why are triangles important when making structures

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

Crazy boy [7]

Answer:

104.3 cm or 179.7

Explanation:

First find time that it takes for the object to hit the ground

$\sqrt{(2H)/g} -> \sqrt{(2 x 179)/ 9.8} = 6.04s\\$ *

Then find xf of projectile $xf= 5.9(6.04) = 37.7\\\\$

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7

hope that helps

4 0

2 years ago

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

(c) $E=2.24*10^{4} N/C$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

2 years ago

A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me

enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.

Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.

Explanation:

6 0

2 years ago

(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

$m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

Now, plug in the given values and solve for 'v'. This gives,

$v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s$

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0

2 years ago