A single beam of light is shone into a glass containing milky water.

You place a 10 kg block on a ramp with an angle of 20 degrees. You push the block up the ramp giving it an initial velocity of 1

Ainat [17]

Answer:

L = 15.97 m

Explanation:

Given:-

- The mass of the block, m = 10 kg

- The inclination of ramp, θ = 20°

- The initial speed, Vi = 15 m/s

- The coefficient of friction u = 0.4

Find:-

find the total distance the block travels before it turns around and slides back down the ramp.

Solution:-

- The total distance travelled by the block up the ramp is defined when all the kinetic energy is converted into potential energy and work is done against the friction. Final velocity V2 = 0.

- Develop a free body diagram of the block. Resolve the weight "W" of the block normal to the surface of ramp. Then apply equilibrium condition for the block in the direction normal to the surface:

N - W*cos( θ ) = 0

Where, N : The contact force between block and ramp.

N = m*g*cos ( θ )

- The friction force (Ff) is defined as:

Ff = u*N

Ff = u*m*g*cos ( θ )

- Apply the work-energy principle for the block which travels a distance of "L" up the ramp:

K.E i = P.E f + Work done against friction

Where, K.E i = 0.5*m*Vi^2

P.E f = m*g*L*sin( θ )

Work done = Ff*L

- Evaluate "L":

0.5*m*Vi^2 = m*g*L*sin( θ ) + u*m*g*cos ( θ )*L

0.5*Vi^2 = g*L*sin( θ ) + u*g*cos ( θ )*L

0.5*Vi^2 = L [ g*sin( θ ) + u*g*cos ( θ ) ]

L = 0.5*Vi^2 / [ g*sin( θ ) + u*g*cos ( θ ) ]

L = 0.5*15^2 / [ 9.81*sin( 20 ) + 0.4*9.81*cos ( 20 ) ]

L = 15.97 m

7 0

3 years ago

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago

Help me answer this

liberstina [14]

no thanks your a fat h/o/e

7 0

3 years ago

The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow

grin007 [14]

Answer:

420m

Explanation:

Given parameters:

Time = 5minutes

Average speed = 1.4m/s

Unknown:

Distance covered = ?

Solution:

Speed is the rate of change of distance with time.

Mathematically;

Speed = $\frac{distance}{time}$

Distance = speed x time

Insert the parameters and solve;

Convert the time to seconds;

1 minute = 60s

5 minute = 5 x 60 = 300s

So,

Insert the parameters and find the distance;

Distance = 300 x 1.4 = 420m

3 0

3 years ago

Which of the following are located outside the nucleus of an atom? Select all that

dmitriy555 [2]

Electrons are located outside the nucleus

3 0

3 years ago