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DedPeter [7]
3 years ago
6

A 4.04 kg block slides down a smooth, frictionless plane having an inclination of 30◦. The acceleration of gravity is 9.8 m/s^2.

The acceleration of the plane is 4.9
What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of 2.37 m down the incline?
Answer in units of N.
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

Explanation:

What a lot of words to solve such a simple problem! The perpendicular force is the one that is pushing straight down on the plane. There is no side to side movement here or friction acting on this dimension at all. Perpendicular force is the same as the weight of the block. That's it! Perpendicular force is also normal force which is the same thing as weight:

w = mg so

w = (4.04)(9.8) and

w = 4.0 × 10¹ N

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The energy from the sun that reaches the corn is about two billionths.
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A 900 N student runs up the stairs 3.5 m high in 12<br> seconds. How much POWER do they generate?
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4 0
3 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
3 years ago
A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What
kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let p_{1} = momentum of the 1st ball

p_{2} = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}

or

(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}

Since we are dealing with identical balls, all the m terms cancel out so we are left with

(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} +  (v_{2})_{f}\cos \theta_{2}

Putting in the numbers, we get

1.33 = (0.750) \cos(33.30)  + (v_{2})_{f} \cos \theta_{2}

=  > (v_{2})_{f} \cos \theta_{2} = 0.703

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}

or

=  > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30)  = 0.412

Taking the ratio of the sine equation to the cosine equation, we get

\frac{ \sin \theta _{2}}{ \cos \theta_{2} }  =  \tan \theta_{2}  =  \frac{0.412}{0.703}  = 0.586

or

\theta_{2}  =  { \tan}^{ - 1} (0.586) = 30.4

Solving now for (v_{2})_{f},

(v_{2})_{f}  =  \frac{0.412}{ \sin(30.4) }  = 0.815 \:  \frac{m}{s}

3 0
3 years ago
The hydrocarbon used to manufacture foam plastics is called styrene. Analysis shows composition to be 92.25% C and 7.75% H. Anal
zhannawk [14.2K]

Answer:

molecular formula = C_{8} H_{8}

Explanation:

Given data

c = 92.25%

H = 7.75%

molar mass = 104 g/mol

to find out

the empirical and molecular formula for styrene

solution

we know that

styrene 1 g contain = 0.9225 g C and 0.0775 g H

so

C = 104 × 0.9225 g / 12 g/mol

C = 7.995 mol = approx 8 mol

and

H = 104 × 0.0775 g / 1 g/mol

H = 8.06 mol = approx 8 mol

so we say that 1 mole of styrene  have 8 mole of C and H

so

molecular formula = C_{8} H_{8}

8 0
3 years ago
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