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Reika [66]
3 years ago
5

A 6 kg block is sliding down a horizontal frictionless surface with a constant speed of 5 m/s. It then slides down a frictionles

s ramp to another horizontal frictionless surface which is 2 m lower than the first horizontal surface. What is the constant speed with which the block is now sliding along this new horizontal surface?
Physics
1 answer:
BARSIC [14]3 years ago
4 0

Answer:

5in

Explanation:

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The airplane is flying with a constant velocity. Which force acting on the airplane below represents the friction from air resis
tankabanditka [31]

The correct answer to the question is  C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically.  There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

8 0
3 years ago
Read 2 more answers
A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v
Masja [62]

Total displacement along the length of mountain is given as

L = 235 m

angle of mountain with horizontal = 35 degree

now we will have horizontal displacement as

x = L cos35

x = 235 cos35 = 192.5 m

similarly for vertical displacement we can say

y = L sin35

y = 235 sin35 = 134.8 m

7 0
3 years ago
Read 2 more answers
What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
VladimirAG [237]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

A

   v_w  =  2.256 *10^{8} \  m/s

B

  v_e  =  2.21 *10^{8} \  m/s

C

The  correct option is  B  

Explanation:

From the question we are told that

      The refractive  index of water is  n_w  =  1.33

      The  refractive  index of ethanol is  n_e  =  1.36

       

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

       v_w  =  \frac{c}{n_w }

Where  c is the speed of light with value  c =  3.0 *10^{8} \  m/s

       v_w  =  \frac{3.0 *10^{8}}{1.33 }

       v_w  =  2.256 *10^{8} \  m/s

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

            v_e  =  \frac{ c}{n_e }

  =>       v_e  =  \frac{3.0 *10^{8}}{1.36 }

=>          v_e  =  2.21 *10^{8} \  m/s

4 0
3 years ago
Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.
atroni [7]

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

Learn more about projectile motion.

brainly.com/question/11049671

#SPJ4

4 0
2 years ago
1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
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