A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller diagram below.
1) Find the total energy for the roller coaster at the initial point.
2) Find the potential energy at point A using the PE formula.
3) Use the conservation of energy to find the kinetic energy (KE) at point B.
4) Find the potential energy at point C.
5) Use the conservation of energy to find the Kinetic Energy (KE) of the roller coaster at point C.
6) Use the Kinetic Energy from C, find velocity of the roller coaster at point C.
Answers
For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the initial point is 19820 J
2) The potential energy at point A is 19620 J
3) The kinetic energy at point B is 10010 J
4) The potential energy at point C is zero
5) The kinetic energy at point C is 19820 J
6) The velocity of the roller coaster at point C is 19.91 m/s
1) The total energy for the roller coaster at the initial point can be found as follows:
[tex] E_{t} = KE_{i} + PE_{i} [/tex]
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The total energy is:
[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]
Hence, the total energy for the roller coaster at the initial point is 19820 J.
2) The potential energy at point A is:
[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]
Then, the potential energy at point A is 19620 J.
3) The kinetic energy at point B is the following:
[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]
[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]
Since
[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]
we have:
[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]
Hence, the kinetic energy at point B is 10010 J.
4) The potential energy at point C is zero because h = 0 meters.
[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]
5) The kinetic energy of the roller coaster at point C is:
[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]
[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]
Therefore, the kinetic energy at point C is 19820 J.
6) The velocity of the roller coaster at point C is given by:
[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]
[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]
Hence, the velocity of the roller coaster at point C is 19.91 m/s.
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Part IV Objects on an incline w/ Tension + Friction
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Answers
(a) The normal force on the sleigh is 596.36 N.
(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.
The given parameters;
mass of the sleigh, m = 63 kginclination of the hill, θ = 15⁰tension on the rein, F = 510 Ncoefficient of friction, μ = 0.25The normal force on the sleigh is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]
The magnitude and direction of acceleration of the sleigh is calculated as follows;
[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]
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The coefficient of kinetic friction between the block and the horizontal surface is 0.051.
The given parameters;
initial velocity, u = 3 m/sdistance traveled by the block, s = 9 mThe acceleration of the block is calculated as follows;
[tex]v^2 = u^2 + 2(-a)s\\\\v^2 = u^2 - 2as\\\\a = \frac{u^2 - v^2}{2s} \\\\a = \frac{3^2 - 0^2}{2\times 9} \\\\a =0.5 \ m/s^2[/tex]
The coefficient of kinetic friction between the block and the horizontal surface is calculated as follows;
[tex]\mu_k mg = ma\\\\\mu_k = \frac{a}{g} \\\\\mu_k = \frac{0.5}{9.8} \\\\\mu_k = 0.051[/tex]
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Alejandro Kirk is the catcher for the Blue Jays’ baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m.
Determine is the acceleration of the ball.
Determine the force applied by Alejandro.
Answers
Answer:
a= -5404.6 [m/s²]; F=785.75 [N].
Explanation:
1. Determine is the acceleration of the ball:
equation of the distance (0.135m) is:
[tex]S=-\frac{at^2}{2} +v_0t,[/tex] where S=0.135[m]; a - required acceleration; t - elapsed time; v₀ - initial velocity (38.2 m/s);
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[tex]\left \{ {{a=-\frac{38.2}{t} } \atop {S=-\frac{at^2}{2}+38.2t }} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {0.5at^2-38.2t=0.135}} \right. \ => \ \left \{ {{a=-\frac{38.2}{t} } \atop {19.1t-38.2t=0.135}} \right. \ => \ \left \{ {{a=-5404.6} \atop {t=0.007}} \right.[/tex]
finally, a≈-5404.6 [m/s²].
2. Determine the force applied by Alejandro.
the energy is:
[tex]E=\frac{mv^2}{2}; \ or E=FS, where \ m-the \ mass; \ v-velocity; \ F-required \ force; S-distance;[/tex]
According to these two equations, the required force is:
[tex]F=\frac{E}{S}=\frac{mv^2}{2S};[/tex]
F=0.145*38.8²/2/0.135≈785.75 [N].
note, the suggested way is not the shortest one and not the only one.
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Answer:
Hey mate....
Explanation:
This is ur answer.....
Green & YellowHope it helps!
Brainliest pls!
Follow me! :)
The ground wire on a household circuit is green in color in a typical household
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The charge on one electron is -1.6 ×10⁻¹⁹ coulomb.
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live, brown
neutral is blue.
stripes with green and yellow: earth.
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Hi there!
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[tex]\boxed {\boxed {\sf 27 \ m/s}}[/tex]
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We will use the following kinematic equation:
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[tex]v_i=[/tex]15 m/s[tex]a=[/tex]3 m/s ²t= 4 sSubstitute the values into the formula.
[tex]v_f= 15 \ m/s + (3 \ m/s^2* 4 \ s)[/tex]
Multiply the numbers in parentheses.
3 m/s/s * 4 s = 3 m/s *4 = 12 m/s[tex]v_f= 15 \ m/s +12 \ m/s[/tex]
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Answer:
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Answer:
False
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It is NOT possible for just three forces to be acting upon an object and they still balance each other. A free-falling object experiences a balance of forces.
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A flea sits on the edge of a classic record album that is making 1 revolution in 0.56s. The flea is 0.20m from the center of the record
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Hi there!
We can begin by finding the angular velocity using dimensional analysis:
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(A)
Find the flea's velocity using the following relationship:
v = ωr
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(B)
Centripetal acceleration is given by:
a = ω²r
a = (11.22²)(0.20) = 25.18 m/s²
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Answers
Answer:
A cannon ball of mass 4.0 kg is fired from a stationary 96 kg cannon at 120 m/s. Calculate the velocity of the cannon immediately after firing.
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Momentum of cannon ball after firing = 4.0 × 120 = 480 kg m/s.
Momentum of cannon after firing = -480 kg m/s (because it recoils in the opposite direction and 480 - 480 = 0 kg m/s, the total momentum after collision).
Rearrange p = m v to find v:
v = p/m
v = 480 divided by 96
v = 5.0 m/s
Note that the forward velocity of the cannon ball was given a positive value. The negative value for the cannon's velocity shows that it moved in the opposite direction.
Any vector can be expressed as the sum of two components, usually
referred to as what?
Answers
Two vectors can be added together to determine the result (or resultant). ... computing the vector sum of all the individual forces acting upon that object.
i hope u understand
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