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belka [17]
3 years ago
12

Two masses sit at the top of two frictionless inclined planes that have different angles,__deleted9917f34947e1359b5705bbac0d9227

f8ec5df862078213a2db19c72f00189109deleted__ 0N86-C1-52-40-A837-22820 50% Part (a) What can be said about the speeds of the two masses at the bottom of their respective paths
Physics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

v = \sqrt{2gh}

the speed in the two planes will be the same since it does not depend on the angle of the same

Explanation:

In this exercise we are told that the two inclined planes have no friction force, so we can apply the conservation of energy for each one, we will assume that the initial height in the two planes is the same

starting point. Highest part of each plane

         Em₀ = U = m g h

final point. Lowest part of each plane

        Em_{f} = K = ½ m v²

as there is no friction, the mechanical energy is preserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = \sqrt{2gh}

As we can see, the speed in the two planes will be the same since it does not depend on the angle of the same

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a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration th
vekshin1

the average acceleration of the bullet through the board   -55657×10⁵ m/s²

acceleration-  Rate of change of velocity with respect to time.

S = 0.10 mm = 10⁻⁵ m  ( distance)          [∵ 1mm = 10⁻³]

u =  480 m/s (initial velocity)

v = 345 m/s (final velocity)

As we know the 3rd equation of motion.  

v² = u² + 2aS

a = ? ( acceleration)

using these values in equation we get

(345)² = (480)² + 2×a× 10⁻⁵

a = (345)² -  (480)² / 2× 10⁻⁵

a =  -55657×10⁵ m/s²

The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.

the average acceleration of the bullet through the board  -55657×10⁵ m/s²

The given question is incomplete. The complete question is.

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.

What is the average acceleration of the bullet through the board/?

to know more about  the equation of motion :

brainly.com/question/13269040

#SPJ4

5 0
2 years ago
What is the formula for Impedance for circuit with R, C, and L?
NikAS [45]

Answer:

 The formula for Impedance for circuit with R, C, and L are:

    z=\sqrt{R^{2}+\left ( \omega L-\frac{1}{\omega C} \right )^2}

The impedance Z of the series RLC circuit depend upon angular frequency ω. Impedance is measured in ohms and resistance (R), inductance reactance and capacitive reactance. Series RLC circuit consist of the resistance, a capacitance and an inductance connected in the circuits. Electrical impedance is the measurement of the opposition that a circuit present in a current.

3 0
3 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
choli [55]

Answer:

a)  F = 1.26 10⁵ N, b)  F = 2.44 10³ N, c)   F_net = 1.82 10³ N  directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

         I = Δp

         F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

         F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

         F = m v₀ / t

         F = 63.5 7.89 / 3.99 10⁻³

         F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

          F = 63.5 7.89 / 0.205

          F = 2.44 10³ N

c) The net force is

         F_net = Sum F

         F_net = F - W

         F_net = F - mg

let's calculate

         F_net = 2.44 10³ - 63.5 9.8

         F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

6 0
3 years ago
If a force of 26 N is exerted on two balls, one with a mass of 0.52 kg and the other with a mass of 0.78 kg, the ball with the m
gogolik [260]
False is the correct answer
6 0
3 years ago
A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back
Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m

So, the stone would travel 1.27 m up

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s

Velocity as the rock passes through the original point is 5 m/s

s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s

Speed of the ball at 15 m above the ground is 17.8542 m/s

s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s

v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229

Coefficient of sliding friction is 0.229

8 0
3 years ago
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