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MrRissso [65]
2 years ago
14

A projectile is launched horizontally east at a speed of 29.4 M/s towards a wall 88.2 m away. What is the velocity of the projec

tile at the moment it strikes the wall?
A. 29.4 m/s down
B. 29.4 m/s east
C. 29.4 m/s at 45(degrees) down from horizontal toward the east
D. 41.6 m/s at 45(degrees) down from horizontal toward the east
Physics
1 answer:
Drupady [299]2 years ago
6 0

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s  (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

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Henri draws a wave that has a 4 cm distance between the midpoint and the trough. Geri draws a wave that has an 8 cm vertical dis
andre [41]

Answer:

Henri’s wave and Geri’s wave have the same amplitude and the same energy

Explanation:

The amplitude of a wave is the distance between the midpoint and the trough (or the crest). This is equivalent to half the distance between the trough and the crest. Therefore:

  • amplitude of Henri's wave: 4 cm
  • amplitude of Geri's wave: 8/2 = 4 cm

The energy of a wave is directly proportional to its amplitude.

6 0
2 years ago
Read 2 more answers
Two resistors of resistances R1 and R2, with R2&gt;R1, are connected to a voltage source with voltage V0. When the resistors are
ehidna [41]

Answer

The Value of  r  = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

                               R_T = R_1 +R_2

 And from Ohm law

                           I_s =\frac{ V}{R_T}

                            I_s  = \frac{V_0}{R_1 +R_2} ---(1)

The mathematical representation of the two resistors connected in parallel  is

                    R_T = \frac{1}{R_1} +\frac{1}{R_2}

                          = \frac{R_1 R_2}{R_1 +R_2}

From the question I_p =10I_s

          =>                 I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} }  = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)

     Dividing equation 2 with equation 1

       =>                 \frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}

                                  10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)

We are told that    r = \frac{R_1}{R_2} \ \ \ \ \  = > R_1 = rR_2

From equation 3  

                            10 = \frac{(1-r)^2}{r}

=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  1+r^2 + 2r = 10r

=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0

Using the quadratic formula

                             r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}

        a = 1  b = -8 c =1  

                              =  \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}

                               r= \frac{8+ \sqrt{60} }{2}  \ or \  r = \frac{8 - \sqrt{60} }{2}

                              r = \ 7.87\ or \  r \  = \ 0.127

Now  r =  0.127 because it is the least value among the obtained values

                               

                                   

                             

4 0
2 years ago
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what
hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
Nisha was making a list of things that are obtained from plants and Amol was making a list of things that can dissolve in water.
Allisa [31]

Answer:

Sugar

Explanation:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

Nisha's list is made up of things that can be derived from plants. From the list given, oil, sugar and cotton can also be obtained from plants.

Most plant materials are organic matter.

Amol's list is made up of things that can dissolve in water. Sugar and table salt can also be dissolved in water.

Water is able to dissolve these materials because they are polar compounds. One rule of solubility is that like dissolves like.

New list:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

  Sugar                                       Sugar

                                   

Only sugar from the list can be added to both lists. It can be obtained from plant and can also dissolve in water.

                                             

6 0
2 years ago
A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t
klasskru [66]

Answer:

a.a. \ \frac{7}{25}

b.\ \ \ P(D_1D_2)=\frac{6}{25}

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let A_i=i^{th} photo is accepted and the probability D_i=i^{th} is defected.

Therefore:

P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0
2 years ago
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