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Katena32 [7]
3 years ago
15

Mbbnmzmzkdjhxhxndmfmfmckcjcjcncncngmgkgjcjcj

Physics
2 answers:
Amiraneli [1.4K]3 years ago
8 0
Gmsisient djaektnfidiaus wjrnrksiauenrnsiaur msiwlwkrjsuakeurjrndkaiakejrjd laismrnff
likoan [24]3 years ago
6 0

Answer:

uxiakqoqkq81i101092ie9endndjdjd djdkpwp1pqmsoxkxoskwnnwkw

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What is the average speed of a train that travels at 100km/hr for 6 hours and then 120km/hr for 5 hours
Sindrei [870]

Answer:

Explanation:

Average speed = Total distance / Total time.

100 km/hr

r = 100 km / hr

t = 6 hours

d = 6 * 100 = 600 km

120 km / hr

r = 120 km / hr

t = 5 hour

d = 120 * 5

d = 600 km

Total distance = 600 + 600 = 1200 km

Total time = 5 hour + 6 hours = 11 hours.

Average speed = 1200 km / 11 hours = 109.1

5 0
3 years ago
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30
dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

4 0
3 years ago
A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest
ZanzabumX [31]
<span>1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.</span>
8 0
3 years ago
Read 2 more answers
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
Liula [17]

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

6 0
3 years ago
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