A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.378 s after leaving the slide. Ignore friction and air resistance, find the height H in the drawing.
Answer:
9.63 m
Explanation:
We know that
Δx = vt.
Therefore, v = Δx / t
v = 5 / 0.378
v = 13.23 m/s
h = ½gt, so that
h = ½ * 9.8 * 0.378²
h = 4.9 * 0.1429
h = 0.7 m
Finally, using the law of conservation of energy, we know that
½mv² = mg(H - h)
½v² = g(H - h)
½v²/g = H - h, substituting the values and solving, we have
(½ * 13.23²) / 9.8 = H - 0.7
(½ * 175) / 9.8 = H - 0.7
87.5 / 9.8 = H - 0.7
8.93 = H - 0.7
H = 8.93 + 0.7
H = 9.63 m
A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 15 000 1 10 000x 2 25 000x2, where x is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If
Answer:
9000 J
Explanation:
weight of bullet = 100 g
length of gun barrel = 0.6m
A) determine the work done by the gas on the bullet as the bullet travels the length of the barrel
The work done is an integral of force over distance in this scenario
lets represent the force using some constants ( arbitrary )
Fx = a + bx + cx^2
where : a = 15000, b = 10000, c = - 25000
hence work done ( W ) = [tex]ax_{f } + \frac{bx^{2} _{f} }{2} + \frac{cx^{3} _{f} }{3}[/tex]
where ; xf = 0.6
hence work done(W)
= 15000(0.6) + [(10000*0.6^2)/2] + [(-25000*0.6^3)/3]
= 9000 J
Which of The following statements best describes the objects motion between zero and four seconds?
Answer:
A OR C
Nearly A.
Explanation:
s this statement true or false?
Hurricanes are classified by three stages in which updrafts of air billow up, updrafts and downdrafts swirl the wind up and down, then downdrafts cause clouds to come apart.
true
false
Answer:
True
Explanation:
A tuna jumps out of the water with an initial velocity of 44 feet per second
Answer:
90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you
Explanation:
plz do so it is true
What happens if atoms lose energy during a change of state?
Answer: they become slower.
Explanation: because they get pulled together.
Answer:
The atoms are pulled together by attractive forces and become more organized.
Explanation:
Atoms lose energy as a gas changes to a solid. ... Atoms gain energy as a solid changes to a liquid. If atoms energy during a change of state, they are pulled together by attractive forces and become more organized.
Reference by Quizlet.
When the momentum of an object increases with respect to time, what is true of the net force acting on it?
a. It is zero, because the net force is equal to the rate of change of the momentum.
b. It is zero, because the net force is equal to the product of the momentum and the time interval.
c. It is nonzero, because the net force is equal to the rate of change of the momentum.
d. It is nonzero, because the net force is equal to the product of the momentum and the time interval.
Answer:
C. It is nonzero, because the net force is equal to the rate of change of the momentum
Explanation:
Momentum of an object is given as;
P = mv = ft
f = mv/t
or f = P/t
where;
P is the momentum of the object
m is mass of the object
v is velocity of the object
f is the applied force on the object
t is time
From the given equation above; the net force acting on the object is equal to the rate of change of the momentum, thus it is non-zero.
Therefore, the correct option is "C"
C. It is nonzero, because the net force is equal to the rate of change of the momentum.
The net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
The problem is based on the impulse-momentum concept. As per the impulse -momentum concept, the magnitude of impulse produced by an object is equal to the change in momentum. Then the expression is,
[tex]I = \Delta P[/tex]
here,
I is the magnitude of impulse. And its value is,
[tex]I = F \times t[/tex]
t is the impact time or time interval.
F is the magnitude of net force.
[tex]\Delta P[/tex] is the change in momentum.
Solving as,
[tex]I= \Delta P\\\\F \times t= \Delta P\\\\F = \dfrac{\Delta P}{t}[/tex]
Clearly, the net force is dependent of the change in momentum with respect to time. So, on increasing the momentum change with respect to time, the net force will also increase with non-zero value.
Thus, we can conclude that the net force acting on the object is non-zero, because the net force is equal to the rate of change of the momentum. Hence, option (c) is correct.
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A piece of lead with a specific heat capacity of 126J/kg°C is given 5000J of energy to heat it from 20°C to 250°C. What was the mass of the piece of lead?
Answer:
m = 0.172 [kg]
Explanation:
To solve this problem we must use the expression of thermal energy that relates the heat to the mass, the specific heat, and the temperature differential.
Q = m*Cp*(T_f - T_i)
where:
m = mass [kg]
Cp = specif heat = 126 [J/kg*C]
Q = 5000 [J]
T_f = final temperature = 250 [C]
T_i = initial temperature = 20 [C]
m = Q/ (Cp*(T_f - T_i))
m = 5000/ (126*(250 - 20))
m = 0.172 [kg]
A big dog B and a small dog S are pulling horizontally in opposite directions on a large bone. The bone’s mass is mmm, the big dog’s leftward force magnitude is F_BF
B
F, start subscript, B, end subscript and the small dog’s rightward force magnitude is F_SF
S
F, start subscript, S, end subscript. The bone accelerates to the right along the table with no friction.
The correct expression for the mass m of the bone is m=Fs-Fb/a.
What is force?The force is defined as the shear stress or pressure applied per unit area.
A big dog B and a small dog S are pulling horizontally in opposite directions on a large bone. The bone’s mass is m, the big dog’s leftward force magnitude is Fb and the small dog’s rightward force magnitude is Fs. The bone accelerates to the right along the table with no friction.
Across the bottom is green grass. In the center, there is a brown table with a large bone on top. To the left is a black outline of a large dog facing rightwards that is pulling leftwards on the bone. To the right is a black outline of a small dog facing leftwards that is pulling rightwards on the bone. The small dog stands on a grey box. Across the bottom is green grass.
Two dogs pulling on a large bone. Big dog and small dog images courtesy of pixabay. Consider rightward as the positive direction.
The net force acting on bone will be
F = Fs -Fb
From the Newton's second law of motion, Force = mass x acceleration
F = ma
m = F/a
m = Fs-Fb/a
Thus, correct expression for the mass m of the bone is m=Fs-Fb/a.
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Dangerous substances in an ecosystem that are harmful to its organisms are:
A sediments. B scavengers. C pollutants. D overpopulation.
Answer:
C pollutants
Explanation:
Pollutants are dangerous substances in the ecosystem that are harmful to its organisms. They are contaminants and effluents that leads to pollution in the environment.
There are different types of pollutants. Most of them are introduced into the environment by anthropogenic activities. Also, some natural events also introduce these pollutants into the environment. These materials pose huge health risks to living organisms. They can render the environment unfit for life.Answer:
c
Explanation:
Two trains travel from New Orleans to Memphis in 5 h. The first train travels at a constant velocity of 80 mph, but the velocity of the second train varies. What was the second train's average velocity during the trip
Answer:
Explanation:
The first train travels at a constant velocity of 80 mph and it takes 5 h to cover the distance so distance between two station = 80 x 5 = 400 m
second train too takes 5 h to cover the same distance so its average velocity
= total distance / total time
= 400 / 5 = 80 m /h
average velocity = 80 m /h
34. If R = 4.0 kΩ, C = 3.0 mF, ε = 15 V, Q = 12 mC, and I = 2.0 mA, what is the potential difference Vb − Va?
Answer:the potential difference Vb − Va = -3 V
Explanation:
The potential difference Vb − Va s given by
Vb − Va= Capacitance / Charge + current x resistance - Electromotive force
Vb − Va=Q/C + I X R - E
V=Q / C=(12 x 10^-3) / (3 x 10^-3)
=0.012/0.003ss
=4V
alsoV= 1R = 2.0 x 10^-3 x 4 x 10^3
=0.002 X 4000
=8
Vb − Va= 4V + 8V- 15V
= -3V
Therefore, the potential difference Vb − Va = -3 V
You are walking in Paris alongside the Eiffel Tower and suddenly a croissant smacks you on the
head and knocks you to the ground. From your handy dandy tourist guidebook, you find that the
height of the Eiffel Tower is 300.5 m. If you neglect air resistance, calculate how many seconds
the croissant dropped before it tagged you on the head.
What is the answer
Convert 200km/h to m/s
Answer:
200 kilometers per hour =
55.556 meters per second
Explanation:
have a great day
Answer:
200km × 1000m/1km = 200000 m
1h × 60 mins/h × 60sec/min = 3600s
now you divide 200000m/3600s
= 55.55m/s
The mass of the moon is about 1/81 the mass of the earth and its radius is 1/4
that of the earth. The
acceleration due to gravity at the earth's surface is 9.8 m/s2. What is the acceleration due to gravity on the moon's surface?
9.8 m/s2
2.8 m/s2
1.9 m/s2
3.0 m/s2
What height does a frictionless playground slide need so that a 35 kg child reaches the bottom at a speed of 6.5 m/s?
Answer:
h = 2.15 m
Explanation:
Given that,
Mass of the child, m = 35 kg
The child reaches the bottom at a speed of 6.5 m/s
We need to find the height slide by a frictionless playground slide. Let it is equal to h. Using the conservation of energy to find it as follows :
[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(6.5)^2}{2\times 9.8}\\\\h=2.15\ m[/tex]
So, it will slide to a height of 2.15 m
The height slid by the frictionless playground can be calculated by equating the kinetic and potential energy equations which is 2.16 meters
Mass of child = 35 kg Velocity, v = 6.5 m/s g = 9.8 m/s²K.E = P.E
K. E = 0.5mv² ; P.E = mgh
0.5mv² = mgh
gh = 0.5v²
h = [(0.5v²) ÷ g]
h = [(0.5 × 6.5²) ÷ 9.8]
h = (21.125 ÷ 9.8)
h = 2.155 m
Therefore, the height slid by a frictionless playground is 2.16 meters
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A 3.0-kg mass is located at (0.0 m, 8.0 m), and a 1.0-kg mass is located at (12 m, 0.0 m) where (x, y) represents x and y coordinates respectively on x-y plane. You want to add a 4.0-kg mass so that the center of mass (or center of gravity) of the three-mass system will be at the origin. What should be the coordinates of the 3.0-kg mass
Answer:
The coordinates of the 4.0-kg mass must be [tex](x,y) = (-3\,m,-6\,m)[/tex].
Explanation:
We have a system formed by three particles with known masses and locations with respect to origin. The location of the center of mass is calculated by means of weighted means. That is:
[tex]\bar x = \frac{\Sigma\limits_{i=1}^{n}\,x_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}[/tex] (Eq. 1)
[tex]\bar y = \frac{\Sigma\limits_{i=1}^{n}\,y_{i}\cdot m_{i}}{\Sigma\limits_{i=1}^{n}m_{i}}[/tex] (Eq. 2)
Where:
[tex]\bar x[/tex], [tex]\bar y[/tex] - Coordinates of the center of mass of the system, measured in meters.
[tex]x_{i}[/tex], [tex]y_{i}[/tex] - Coordinates of the i-th particle, measured in meters.
[tex]m_{i}[/tex] - Mass of the i-th particle, measured in kilograms.
If we know that [tex]m_{1} = 3\,kg[/tex], [tex]m_{2} = 1\,kg[/tex], [tex]m_{3} = 4\,kg[/tex], [tex]x_{1} = 0\,m[/tex], [tex]x_{2} = 12\,m[/tex], [tex]\bar x = 0\,m[/tex], [tex]y_{1} = 8\,m[/tex], [tex]y_{2} = 0\,m[/tex] and [tex]\bar y = 0\,m[/tex], the location of the 4.0 kg mass is:
[tex]\frac{(3\,kg)\cdot (0\,m)+(1\,kg)\cdot (12\,m)+(4\,kg)\cdot x_{3}}{3\,kg+1\,kg+4\,kg} = 0\,m[/tex]
[tex]12+4\cdot x_{3} = 0[/tex]
[tex]x_{3} = -3\,m[/tex]
[tex]\frac{(3\,kg)\cdot (8\,m)+(1\,kg)\cdot (0\,m)+(4\,kg)\cdot y_{3}}{3\,kg+1\,kg+4\,kg}= 0\,m[/tex]
[tex]24+4\cdot y_{3} = 0[/tex]
[tex]y_{3} = -6\,m[/tex]
The coordinates of the 4.0-kg mass must be [tex](x,y) = (-3\,m,-6\,m)[/tex].
The coordinates of the 4kg mass so that the center of mass of the three-mass system will be at the origin is (-3,-6)
Finding the coordinates:The coordinates of the 3kg mass are (0.0 m, 8.0 m)
and the coordinates for the 1kg mass are (12 m, 0.0 m)
Let the coordinates of the 4kg mass be (x,y)
The final coordinates of the center of mass of the system are (0,0),
so the equation for x-coordinate is given by:
[tex]0=\frac{3\times0+1\times12+4\times x}{3+1+4}[/tex]
12 + 4x = 0
x = -3
the equation for y-coordinate is given by:
[tex]0=\frac{3\times8+1\times0+4\times y}{3+1+4}[/tex]
24 + 4y = 0
y = -6
So, the coordinates of the 4kg mass are (-3,-6)
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What is the most logical explanation for why kilo is a more important prefix to remember than tera
A quarterback applies a force 35 N for .28 seconds. What is the
impulse given to the ball?
A 0.150-kg baseball traveling in a horizontal direction with a speed of 21.0 m/s hits a bat and is popped straight up with a speed of 15.0 m/s. If the bat was in contact with the ball for 51.0 ms, what was the average force of the bat on the ball
Answer:
17.65N
Explanation:
We know that the relationship for the impulse and momentum is given as
Ft=mv
given data
mass m=0.15kg
t= 51ms= 26.5/1000= 0.051seconds
velocity v1= 21m/s
Velocity v2=15m/s
substituting in the expression we have
Ft=m(v1-v2)
make F subject of formula we have
F=m(v1-v2) /t
F=0.15(21-15)/0.051
F=0.15*6/0.051
F=0.9/0.051
F=17.65N
This question involves the concepts of momentum and impulse.
The average force of the bat on ball is "17.65 N".
What is the average force?According to Impulse Equation the change in momentum is equal to the product of applied force and time of contact. Mathematically:
[tex]\Delta P=F\Delta t\\\\F=\frac{m\Delta v}{\Delta t}[/tex]
where,
F = Average force = ?m = mass of baseball = 0.15 kgΔv = change in speed = 21 m/s - 15 m/s = 6 m/sΔt = time of contact = 51 ms = 0.051 sTherefore,
[tex]F=\frac{(0.15\ kg)(6\ m/s)}{0.051\ s}[/tex]
F = 17.65 N
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Jack drops a stone from rest off of the top of a bridge that is 23.4 m above the ground. After the stone falls 7.2 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock
Explanation:dropped stone hits water:
24.4-4.9t^2=0
t=2.23
stone has fallen 6.6m:
4.9t^2 = 6.6
t = 1.16
So, Jill's thrown stone only has 2.23-1.16=1.07 seconds to hit the water:
24+1.07v-4.9*1.07^2 = 0
v = -17.19 m/s
Both the rocks hit the water at the exact same time, and the initial velocity of Jill's rock will be -19.28 m/s.
What is velocity?A vector measurement of the rate and direction of motion is what is meant by the term "velocity." Simply said, velocity is the rate of movement in a single direction. Velocity may be used to gauge both the speed of a rocket launching into space and the speed of a car travelling north on a busy freeway.
According to the question, initial velocity will be 0.
By using equation of motion :
Let Jill is x and Jack is y.
s=ut+1/2at² where,
a is acceleration,
u is initial velocity and,
t is the time period.
23.4=(-1/2)(9.8)×t²
t(y)=[tex]\sqrt{(23.4)/(4.9)}[/tex]
=2.185 seconds.
Time taken by Jack's stone that it travel the first 7.2 meters.
t=[tex]\sqrt{(2*7.2)/9.8}[/tex] =1.21 seconds.
So, time taken by stone thrown by jack,
t(Jill)=t(jack)-t
=23.4=-ut(x)-1/2(gt²)(y)
-23.4=(-u×0.975)-(1/2×9.8×(0.975)²)
u= -(23.4-4.6/0.975)
u=-19.28 m/s
Hence, the initial velocity of the Jill's stone will be -19.28 m/s.
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If we put a flammable material in a point where many light rays converge we can set it on fire. With what type of spherical mirror would it be possible to do this
Answer:
Concave mirror
Explanation:
A concave mirror is a type that has the property of converging rays of light that are parallel to the principal axis to a point of focus. Thus, for the convergence of many rays of light (i.e a beam of light) as required in the given question, a concave mirror would serve the purpose. Curved mirrors have various applications in optics.
A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision
Answer:
v₂f = 0.16 m/s (to the right)
Explanation:
Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.The initial momentum is due to the small cart only, because the big one is at rest.Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :[tex]p_{o} = m_{1} * v_{o1} = 0.3 kg * 1.3 m/s (1)[/tex]
The final momentum can be written as follows:[tex]p_{f} = m_{1} * v_{1f} + m_{2} * v_{2f} = 0.3 kg* (-0.86m/s) + 4.00 kg* v_{2f} (2)[/tex]
As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:
[tex]v_{2f} = \frac{0.3kg* (1.3 + 0.86) m/s}{4.00kg} = \frac{0.3kg*2.16m/s}{4.00kg} = 0.16 m/s[/tex]
As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.The tendency of an object to resist change in its motion is known as
a. mass
b. inertia
c. force
d. balance
If the rock is kicked at initial velocity of 12 m/s from a height of 106 m above the ground what horizontal distance does the rock travel before striking the ground?
Answer:
74 m.
Explanation:
From the question given above, the following data were obtained:
Height (h) = 186 m
Initial velocity (u) = 12 m/s
Horizontal distance (d) =?
Next, we shall determine the time taken for the rock to get to the ground. This can be obtained as follow:
Height (h) = 186 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
H = ½gt²
186 = ½ × 9.8 × t²
186 = 4.9 × t²
Divide both side by 4.9
t² = 186/4.9
Take the square root of both side
t = √(186/4.9)
t = 6.16 s
Thus, the time taken for the rock to get to the ground is 6.16 s
Finally, we shall determine the horizontal distance travelled by the rock as follow:
Initial velocity (u) = 12 m/s
Time (t) = 6.16 s
Horizontal distance (d) =?
Horizontal distance (d) = Initial velocity (u) × Time (t)
d = u × t
d = 12 × 6.16
d = 73.92 ≈ 74 m
Therefore, the horizontal distance travelled by the rock is 74 m
119) You throw a rock horizontally off a cliff with a speed of 20 m/s and no significant air resistance. After 2.0 s, the magnitude of the velocity of the rock is closest to
Answer: 44.54 m/s
Explanation:
Let's break the problem into horizontal and vertical.
Horizontal:
Here we do not have any force, only a constant horizontal velocity, so we can write:
Vx(t) = 20m/s.
Vertical:
Here we have the gravitational acceleration acting on the rock, then we can write:
Ay(t) = -9.8m/s^2
For the vertical velocity, we need to integrate over time and get:
Vy(t) = (-9.8m/s^2)*t + V0
Where V0 is the initial vertical velocity, but the rock was thrown horizontally, so we do not have an initial velocity in the vertical axis.
Vy(t) = (-9.8m/s^2)*t.
Ok, now we want to know the magnitude of the velocity when t = 2.0s.
The magnitude will be equal to:
V = √( Vx(2s)^2 + Vy(2s)^2)
= √( (20m/s)^2 + (-2s*9.8m/s^2)^2) = 44.54 m/s
The magnitude of the velocity after 2 seconds is around 44.54 m/s
a decompression chamber used by deep sea divers has a volume of 10.1m^3 and operates at an internal pressure of 4.25 atm. What volume, in cubic meters would the air in the chamber occupy if it were at 1.00 atm pressure, assuming no tempreature change.
Answer:
V2=42.93cm3
Explanation:
According to Gay-Lussac's Law: "The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up."
P1V1=P2V2
Given data
P1=4.25atm
V1=10.1cm³
P2=1atm
V2=?
V2=P1V1/P2
V2=(4.25*10.1)/1
V2=42.93cm3
Which of the following is correct about a motion diagram?
a. A group of images for different objects found in one place at the same time
b. A series of images that show the position of a moving object at equal time intervals
c. A series of images that show the position of a moving object at unequal time intervals
d. Collection of photos taken by a person while moving around different places
Answer:
d. collection of photos taken by a person while moving around different places
Answer:
Moving object at equal time intervals
Calculate the linear acceleration of a car, the 0.300-m radius tires of which have an angular acceleration of 17.0 rad/s2. Assume no slippage and give your answer in m/s2.
Answer:
5.1m/s^2Explanation:
In this problem we want to solve for the linear acceleration
Given data
radius r= 0.3m
angular acceleration α=17rad/s^2
we know that the expression relating angular acceleration and linear acceleration is given as
a=rα
Substituting we have
a=0.3*17
a=5.1m/s^2
A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable
Answer:
425.1 W
Explanation:
We are given;
Counter mass of elevator; m_c = 940 kg
Cab mass of elevator; m_d = 1200 kg
Distance from rest upwards; d = 35 m
Time to cover distance; t = 3.5 min
Now, this elevator will have 3 forces acting on it namely;
Force due to the counter weight of the elevator; F_c
Force due to the cab weight on the elevator; F_d
Force exerted by the motor; F_m
Now, from Newton's 2nd law of motion,
The force exerted by the motor on the elevator can be given by the relationship;
F_m = F_d - F_c
Now,
F_d = m_d × g
F_d = 1200 × 9.81
F_d = 11772 N
F_c = m_c × g
F_c = 940 × 9.81
F_c = 9221.4 N
Thus;
F_m = 11772 - 9221.4
F_m = 2550.6 N
Now, the average power required of the force the motor exerts on the cab via the cable is given by;
P_m = F_m × v
Where v is the velocity of the elevator.
The velocity is calculated from;
v = distance/time
v = 35/3.5
v = 10 m/min
Converting to m/s gives;
v = 10/60 m/s = 1/6 m/s
Thus;
P_m = 2550.6 × 1/6
P_m = 425.1 W