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Anna71 [15]
3 years ago
12

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp

eed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

ipn [44]3 years ago
6 0

Answer:

Explanation:want answers

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What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze
yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of 62.5 cm of H_2O

i.e. h=62.5 cm =0.625 m

Effective area A=51 cm^2

initial Pressure= 1 atm=101.325 kPa

Gauge Pressure P=\rho gh

\rho =density\ of\ water =1000 kg/m^3

P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa

Force creates a pressure of P_1 which will be equal to Gauge Pressure

P_1=\frac{F}{A}

P_1=P_{gauge}

\frac{F}{A}=5.937 kPa

F=5.937\times 51\times 10^{-4}\times 10^3

F=30.27 N

6 0
3 years ago
A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
3 years ago
Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a
Verizon [17]

Answer:

Allison is probably looking at the asteroid.

Explanation:

  • Asteroids are the giant gas balls of inner solar system that are present in the space.
  • Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
  • This is also the reason why it's called as the minor planets.
  • It could have been comet but if it had been a comets he must have seen the tail but asteroids do not have any tail like structure.
8 0
3 years ago
The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how
liraira [26]

Answer:

A

   x = 0.198456 \ m

B

    h  =  1.3061 \  m

C

 v =  5.06 \  m/s

D

  d = 4.0273 \  m

Explanation:

Considering the first question

From the question we are told that

   The spring constant is  k  =  32.50 N/m

    The potential energy is  PE  =  0.640 \ J

Generally the potential  energy stored in spring  is mathematically represented as   PE  =  \frac{1}{2}  *  k  *  x^2

=>    0.640=  \frac{1}{2}  * 32.50  *  x^2  

=>    x = \sqrt{0.03938}  

=>    x = 0.198456 \ m  

Considering the second question

 From the question we are told that

   The mass of the dart is  m =  0.050 kg

Generally from the law of energy conservation

         PE =  mgh

=>       0.640   =  0.050 *  9.8  *  h

=>      h  =  1.3061 \  m

Considering the third  question

   The height at which the dart was fired horizontally is  H  =   3.90\  m

Generally  from the law of energy conservation

         PE = KE

Here  KE is kinetic energy of the dart which is mathematical represented as

     KE  =  \frac{1}{2}  *  mv^2

=>      0.640 =  \frac{1}{2}  * 0.050 *  v^2

=>       v^2 = 25.6

=>       v =  5.06 \  m/s

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

       t  =  \frac{ 2 *  H }{g}

=>     t  =  \frac{ 2 * 3.90 }{9.8 }

=>     t  =  0.7959 \ s

Generally the  horizontal distance from the equilibrium position to the ground is  mathematically represented as

       d =  v  *   t

=>     d = 5.06  *   0.7959

=>     d = 4.0273 \  m

5 0
2 years ago
A rocket is launched from the launch pad at a speed of 700 km/hr.
Juliette [100K]

Answer:

Explanation:

Vm = Δs/Δt

700km/h = Δs/1.5h

700 = Δs/1.5

\frac{700}{1} = \frac{S}{1.5}

S = 700 x 1.5

S = 1050 Km

*(S = Δs)

Answer: The rocket will have traveled 1050 Km

Hope this help ☺

5 0
3 years ago
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