Answer:
945 m
Explanation:
First of all let's find the first acceleration before the first stop. We will use Newton's 3rd equation of motion;
v² = u² + 2as
a = (v² - u²)/2s
In this first motion, the initial velocity is given as u = 67 m/s, while the final velocity which is the stop is v = 0 m/s and the distance is s = 250 m
Thus;
a = (0² - 67²)/(2 × 250)
a = -4489/500
a = -8.978 m/s²
Now, to get the time spent for this 250 m distance, we will use Newton's 2nd law of motion;
s = ut + ½at²
Plugging in the relevant values;
250 = 67t + ½(-8.978)t²
250 = 67t - 4.489t²
4.489t² - 67t + 250 = 0
Using quadratic formula, we will arrive at;
t = 7.463 s
Now, we are told that she accelerates out, reaching her previous speed of 67.0 m/s after a distance of 360 m. Thus, let's calculate the time it took to accelerate after a distance of 360 m.
Since she stayed from rest in this phase, then initial velocity (u) = 0 m/s.
Like in the first phase, we can find the acceleration from;
a = (v² - u²)/2s
a = (67² - 0²)/(2 × 360)
a = 4489/720
a = 6.235 m/s²
Thus, we can find the time for this 360 m phase by using; s = ut + ½at², we have;
360 = 0 + ½(6.235)t²
Multiply through by 2 to get;
6.235t² = 720
t² = 720/6.235
t² = 115.477145
t = √115.477145
t = 10.746 s
We are told that she spends 5 s in the pit. Thus;
Total time is;
t_total = 7.463 + 5 + 10.746
t_total = 23.209 s
Distance covered by the Mercedes Benz = 67 × 23.209 = 1555 m
Total distance covered by Thunderbird = 250 + 360 = 610 m
Thus, distance of Thunderbird behind Mercedes Benz is = 1555 - 610 = 945 m
An athlete at the gym holds a 3.5kg steel ball in his hand. His arm is 70cm long and has a mass of 4.0kg.What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?What is the magnitude of the torque about his shoulder if he holds his arm straight, but 55 degrees below horizontal?
Answer:
A. 38 nm
B. 22 nm
Explanation:
This question has given us m, mass of the ball as 3.5kg
M = The mass of arms = 4kg
R = length of arms as 70cm
70 cm = 0.7m
0.7m/2 = 0.35m
rmg + Rmg
g = 9.8m/s
= 0.7(3.5)(9.8)+(0.35)(4)(9.8)
= 24.01 + 13.72
= 37.72
~38
B.
55 degrees below horizontal
Cos 55⁰ = 0.5736
= 0.7(3.5)(9.8)(0.5736)+(0.35)(4)(9.8)(0.5736)
= 13.772136+7.869792
= 21.642
~22Nm
The magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
The formula for calculating the magnitude of the torque about his arm is expressed as:
[tex]\tau = (F_1r + F_2R) cos\theta[/tex]
F1 and F2 are the forces
[tex]\theta[/tex] is the given angle
F1 = 3.5 * 9.8 = 34.3 N
r = 70cm = 0.7m
F2 = 4.0 * 9.8 = 39.2N
R = 0.7/2 = 0.35m
Substitute the given values into the formula:
[tex]\tau = (34.3(0.7) + (39.2)(0.35)) cos55^0\\\tau = (24.01+13.72)cos55\\\tau = 37.73cos55\\\tau=21.64Nm[/tex]
Hence the magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm
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In a popular demonstration, vinegar and baking soda are combined in a bottle as shown. Sodium carbonate, water as well as carbon dioxide gas are formed. The balloon traps the carbon dioxide.
According to the law of conservation of mass, the total mass of the reactants equals the total mass of the products. How could you alter the demonstration above so that the mass of products would not equal the total mass of the reactants?
A) Use less of the reactants.
B) Do not use a clean bottle.
C) Take the balloon off the bottle.
D) Shake the bottle while the chemicals are reacting.
Answer:
Shake the bottle while the chemicals are reacting.
Explanation:
Answer:
C
Explanation:
If you take the balloon off, the product will leak.
2. An aluminum plate has a circular hole cut in it. A copper ball (solid sphere) has exactly the same diameter as the hole when both are at room temperature, and hence can just barely be pushed through it. If both the plate and the ball are now heated up to a few hundred degrees, the ball will fit more easily. Why
Answer:
Thermal expansion
Explanation:
One of the effects of heat is that it causes expansion of materials.
When the copper ball and aluminium plate are both heated, the both materials expand. When they expand, the copper ball can is found to fit more easily into the aluminium plate after both metals have become wider than they were at room temperature.
The coefficient of thermal expansion for aluminium is 35% greater than that of copper. This is the reason why the copper ball can fit into the aluminum plate more easily at higher temperature. The aluminum has expanded 35% more than copper making it easier and leaving a lot of space for the copper to fit into the aluminum plate.
A 54.0-kg ice skater is moving at 3.94 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.900 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. .931416 Correct: Your answer is correct. kN (b) Compare this force with her weight. Frope W
Answer:
Explanation:
a ) magnitude of force = centripetal force = m v² / R , m is mass , v is velocity of skater , R is radius of circular path .
= 54 x 3.94² / .9
= 931.41 N
= .93141 kN.
b ) Her weight = mg = 54 x 9.8 = 529.2 N
= .5292 kN .
Ratio = .93141 / .5292 = 1.76
A car going around a curve of radius R at a speed V experiences a centripetal acceleration ac. What should be the radius of the path that leads to same centripetal acceleration while it travels at speed 2V
Answer:
4R
Explanation:
The centripetal acceleration a is given by a = v²/r where V = speed of object and r = radius of circle. Now, it is given that V = speed of car = V and radius of circle = R, so its centripetal acceleration a = V²/R.
Now, if the speed is doubled to V' = 2V with the same centripetal acceleration, we find the new radius of the curve R' from
a = V'²/R'
R' = V'²/a
R' = (2V)² ÷ V²/R
R' = 4V²R/V²
R' = 4R
So, the new radius is 4 times the initial radius.
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F, what would be the volume of air in ft³ exiting the evaporator coil?
Given :
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.
To Find :
The volume of air in ft³ exiting the evaporator coil.
Solution :
We know, relation between initial and final temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
Here, temperature is in Kelvin .
We know, Kelvin temperature is given by :
[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]
Putting F = 60° F,
[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]
Putting F = 80° F,
[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]
Putting all values, we get :
[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]
Hence, this is the required solution.
Very Urgent!!!!!!! Please!!!!!!!!
A 35kg child standing on ice (frictionless) pushes a 7kg bowling ball with a force of 100N. What is the force acting on the 35kg child? What is the acceleration of the 35kg child?
Please real answers!! Gives many points!!!!!!
Answer:
a = 2.85714 m/s^2
Explanation:
Fnet=ma
eaqual and oppisate forces means that the 100N pushed on the ball comes back to the child.
100 = 35*a
a= 100/35
frictionless floor by force . What total mass is accelerated to the right by (a) force , (b) cord 3, and (c) cord 1
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]F = 20 \ kg[/tex]
b
[tex]F = 18 \ kg[/tex]
c
[tex]F = 10 \ kg[/tex]
Explanation:
From the diagram we are told that
The first mass is [tex]m_1 = 10 \ kg[/tex]
The second mass is [tex]m_2 = 3 \ kg[/tex]
The third mass is [tex]m_3 = 5 \ kg[/tex]
The fourth mass is [tex]m_4 = 2 \ kg[/tex]
Gnerally the total mass accelerated by right force is mathematically evaluated as
[tex]F = m_1 + m_2 +m_3 + m_4[/tex]
[tex]F = 10 + 3 +5 + 2[/tex]
[tex]F = 20 \ kg[/tex]
Gnerally the total mass accelerated by cord 3 is mathematically evaluated as
[tex]F = m_1 + m_2 +m_3[/tex]
[tex]F = 10 + 3 +5[/tex]
[tex]F = 18 \ kg[/tex]
Gnerally the total mass accelerated by cord 3 is mathematically evaluated as
[tex]F = m_1[/tex]
[tex]F = 10 \ kg[/tex]
1. A balloon drifts 100 m towards the west in 10 seconds, then the wind suddenly changes
and the balloon flies 60m towards the east in the next 6 seconds.
a) What total distance did it travel? 160 meters
b) What was the average speed during the first 10 seconds?
c) What was its average speed during the next 6 seconds?
d) What was its average speed for the entire trip?
e) What was its total displacement?
1) What was its average velocity during the first 10 seconds?
g) What was its average velocity during the next 6 seconds?
h) What was its average velocity for the entire trip?
Answer:
Explanation:
a) The total distance traveled by the balloon = (100 + 60) m
= 160 metes
b) speed = [tex]\frac{distance covered}{time taken}[/tex]
The average speed for the first flight = [tex]\frac{100}{10}[/tex]
= 10 m/s
c) The speed during its next flight = [tex]\frac{60}{6}[/tex]
= 10 m/s
d) Average speed for the entire trip = [tex]\frac{(10 + 10)}{2}[/tex]
= 10 m/s
e) Total displacement = [tex]\sqrt{(100)^{2} + (60)^{2} }[/tex]
= [tex]\sqrt{13600}[/tex]
= 116.62 m
f) velocity = [tex]\frac{displacement}{time}[/tex]
Average velocity during the first 10 seconds = [tex]\frac{100}{10}[/tex]
= 10 m/s
g) Average velocity during the next 6 seconds = [tex]\frac{60}{6}[/tex]
= 10 m/s
h) Average velocity for the entire trip = [tex]\sqrt{(10)^{2} + (10)^{2} }[/tex]
= [tex]\sqrt{200}[/tex]
= 14.14 m/s
Starting at rest at the top of a ski slope, you travel down the hill reaching a velocity of 20 m/s after 13 seconds. Once you reach the bottom of the hill you travel at a constant velocity of 20 m/s for the next 7 seconds. Sketch a quantitative (with numbers) position vs time, velocity vs time, and acceleration vs time graph of the skier's motion. PLEASE SHOW WORK AND I WILL MARK YOU THE BRAINLIEST
Answer:
The graphs are attached
Explanation:
We are told that he started at rest at rest and travelled down the hill reaching a velocity of 20 m/s after 13 seconds.
Acceleration is gotten from;
v = u + at
a = (v - u)/t
a = (20 - 0)/13
a = 20/13 m/s² or 1.54 m/s²
Distance in this period is gotten from;
v² = u² + 2as
s = (v² - u²)/2s
s = (20² - 0²)/(2 × 20/13)
s = 400/(40/13)
s = 130 m
We are told that after reaching the bottom of the hill, he travelled at a constant velocity of 20 m/s for the next 7 seconds.
At constant velocity, acceleration is 0.
Thus,distance in this period is;
s1 = vt = 20 × 7 = 140 m
I've attached the graphs
An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball
Answer:
v = 3.951 m/s
Explanation:
Given that,
Mass of a ball, m = 6.5 kg
Radius of the circle, r = 0.9 m
Angular speed of the ball, [tex]\omega=0.7\ rev/s=4.39\ rad/s[/tex]
Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :
[tex]v=r\omega\\\\v=0.9\times 4.39\\\\v=3.951\ m/s[/tex]
So, the tangential speed of the ball is 3.951 m/s.
An object thrown straight upward reaches a height of 10 meters. If the object is thrown upward with twice this speed, it will reach a height of ____ m.
Answer:
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
Explanation:
Based on the Principle of Energy Conservation we find that the square of the initial speed of the object ([tex]v[/tex]), measured in meters per second, is directly proportional to its change in height ([tex]\Delta h[/tex]), measured in meters. That is:
[tex]v^{2} \propto \Delta h[/tex]
[tex]v^{2} = k\cdot \Delta h[/tex] (Eq. 1)
Where [tex]k[/tex] is the proportionality constant, measured in meters per square second.
Such constant is eliminated by using the following relationship:
[tex]\left(\frac{v_{2}}{v_{1}} \right)^{2} = \frac{\Delta h_{2}}{\Delta h_{1}}[/tex]
[tex]\Delta h_{2} = \left(\frac{v_{2}}{v_{1}} \right)^{2}\cdot \Delta h_{1}[/tex] (Eq. 2)
Where:
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the object, measured in meters per second.
[tex]\Delta h_{1}[/tex], [tex]\Delta h_{2}[/tex] - Initial and final changes in height, measured in meters.
If we know that [tex]\frac{v_{2}}{v_{1}} = 2[/tex] and [tex]\Delta h_{1} = 10\,m[/tex], then the change in height is:
[tex]\Delta h_{2} = 2^{2}\cdot (10\,m)[/tex]
[tex]\Delta h_{2} = 40\,m[/tex]
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
When the object is thrown with double the speed, it reaches a height of 40m.
Laws of motion:Given that the height attained by the object is h = 10m.
Let the initial speed of the object be u
and its final speed will be v = 0 at the highest point.
Then from the third equation of motion, we get,
[tex]v^2=u^2-2gh\\\\0=u^2-2gh\\\\h=\frac{u^2}{2g}[/tex]
if the object is thrown at double the speed, that is u' = 2u, then the height attained will be:
[tex]h'=\frac{u'^2}{2g}\\\\h'=\frac{(2u)^2}{2g}\\\\h'=\frac{4u^2}{2g}\\\\h'=4h[/tex]
h' = 4 × 10m
h' = 40m
So, the height attained will be 40 m.
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This is the motion of an object when the only force acting on it is the force of gravity.
A) Newton's Third Law
B)Newton's First Law
C)Free fall
D)Terminal velocity
Answer:
c it is free fall when an object force act on the force of gravity
how much heat is needed to convert 1.0kg of ice at -10°C to steam at 100°C?
Q= Q1 +Q2 +Q3+Q4+ Q5
3066 J
Why is force a vector
Answer:
A force vector is a representation of a force that has both magnitude and direction.
Car is going from 0 to 60 miles per hour in 6 seconds what is the advertised acceleration in mph squared
Answer:
10 m/h^2
Explanation:
a=v/t
a=60/6
a=10 m/h^2
4. From home a car drives 16 km east and then 24 km south. What is the resultant vector for
the car?
Which is the dependent variable and what is the trend in the graph?
Experimental Solubility Data for a Sugar
140
130
120
110
Solubility (g/100g H,0)
100
90
80
0
10
50
60
20 30 40
Temperature (*C)
Solubility; decreasing
Solubility; increasing
Temperature; decreasing
Temperature; increasing
Answer: Solubility; decreasing
Explanation:
A 4.33-kg soccer ball rolling eastward at a speed of 2.74 m/s is kicked so that it reverses direction and attains a speed of 6.35 m/s. If the duration of the interaction is 55.33 ms, what is the average force on the ball by the player's foot, in N
Answer:
0.71 N
Explanation:
The inertia of a body, I, is the product of the force applied on it and the time which it acts. Also, it is equal to the change in the momentum, P, of a body.
So that:
I = Ft
I = ΔP
Ft = m(v - u)
Where F is the force, t is the time, m is the mass of object, v is the final velocity and u is the initial velocity.
Given that: m = 4.33 kg, u = -2.74 m/s, v = 6.35 m/s and t = 55.33 s
F x 55.33 = 4.33(6.35 - (-2.74))
F x 55.33 = 4.33(6.35 + 2.74)
55.33 F = 4.33 x 9.09
55.33 F = 39.3597
F = [tex]\frac{39.3597}{55.33}[/tex]
= 0.711363
The average force on the ball by the player is 0.71 N.
A dog with a mass of 57.0 kg slides down a wet slope with negligible friction. The dog starts from rest and has a speed of 2.10 m/s at the bottom. What is the height of the slope (in m)?
The dog starts from rest and has a speed of 2.10 m/s at the bottom. The height of the slope will be 0.225 m
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.
As there is almost zero friction between the total potential energy height is equal to the kinetic energy
m*g*h = 1/2* m*v²
Where m is the mass of the dog
g is the acceleration due to gravity,h is the height of the slope
v is the velocity of the sliding dog
g*h = 1/2* v²
9.81*h = 0.5* 2.1²
=0.225 m
The height of the slope would be 0.225 m.
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Determine the xx component of force at point CC in the pipe assembly. Neglect the weight of the pipe. Take F1F1
Answer:
Hello your question has some missing part and diagram attached below is the missing part and diagram
Determine the x component of force at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = { 348 i - 427 j} lb and F2 = {-307 j + 122 k} lb.
Determine the y component of the force at point C in the pipe assembly.
Determine the x component of the force at point C in the pipe assembly.
Determine the x component of the moment at point C in the pipe assembly.
Determine the y component of the moment at point C in the pipe assembly.
Determine the z component of the moment at point C in the pipe assembly.
answer: attached below is the detailed solution
Explanation:
Attached below is the detailed solution
A cat is running at 8.0 m/s and trying to catch a mouse, who has a tenth of the mass of the cat. If the cat speeds up by 2.0 m/s, the cat has ten times of the kinetic energy of the mouse. What is the speed of the mouse
Answer:
The speed of the mouse is 10 m/s
Explanation:
Given;
speed of the cat, u = 8 m/s
let the mass of the cat = m
then, mass of the mouse, = m / 10
let the speed of the mouse = v
Let the kinetic energy of the mouse = K.E
When the cat speeds up by 2 m/s, its new speed = 10 m/s,
then its kinetic energy is given as;
¹/₂m(10)² = 10K.E
[tex]50 m = 10[\frac{1}{2} *\frac{m}{10}.v^2][/tex]
50 = ¹/₂v²
v² = 100
v = √100
v = 10 m/s
Therefore, the speed of the mouse is 10 m/s
A ball rolls at a speed of 4 m/s off a flat surface which is a height 1.3m above the ground. how long (in seconds) is the ball in the air before it hits the ground
34)
Elements are composed of
A)
many type of atoms.
B)
the same type of atom.
C)
many different compounds.
D)
the same type of compound.
What is conserved in physical changes?
O shape
O energy
O mass
Odensity
True or false: Electromagnetic radiation cannot travel through a vacuum.
A
True
B
False
Answer: that one is wrong i did it on flocabulary its false
Explanation:
Since electromagnetic radiations do not require a material medium for their propagation and can travel through vacuum, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
What is an electromagnetic radiation?!An electromagnetic radiation is a radiation which results from.the interaction of the electric and magnetic fields.
Electromagnetic radiations do not require a material medium for their propagation, and hence can travel through vacuum.
An example of electromagnetic radiation is radio waves.
Hence, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.
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In 2 to 3 sentences explain the roles of chloroplast and mitochondria in the process of plant cells gaining energy.
Answer:
Explanation:
Mitochondria is the powerhouse of the cell. And Chroloplast helps plants capture energy
The earth travels at 68,000 miles/hour as it moved around the sun. How many miles does the earth travel in one trip around the sun
Answer:
595,680,000 miles
(= 5.96 x 10⁸ miles)
Explanation:
we are given the speed of travel as 68,000 mi/hr
if we assume one common year
= 365 days
= 365 days x 24 hours/day
= 8760 hours
Then the distance travelled
= speed x time
= 68,000 x 8760
= 595,680,000 miles
= 5.96 x 10⁸ miles (rounded to 3 sig. fig and expressed in scientific form)
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Hope this helps!
A jogger runs at a speed of 4 m/s for 500]s, slows to 2.5 m/s for the next
300 s and then travels the final 900 s at a speed of 3.5 m/s. What is the
average speed of the jogger in m/s?
O 10 m/s
O 3.47 m/s
O 5 m/s
O 3.33 m/s
Answer:
3.33 m/s
Explanation:
Just add up all of the speeds and divide it by 3 (since there are 3 different speeds)
(4 m/s + 2.5 m/s + 3.5 m/s) / 3 = 3.33 m/s
A person throws a shot put from height of 5 feet with an initial vertical velocity of 40 feet per second. Use the vertical motion model, h = -16t^2 + vt + s, where v is the initial velocity in feet per second and s is the initial height in feet, to calculate the amount of time the shot put is in the air before it hits the ground.
Answer:2.6
Explanation: