Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g [tex]\frac{m_1}{m_1 + m_2}[/tex]
let's calculate
a = 0.65 9.8 [tex]\frac{2000}{2000+8000}[/tex]
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
Which letter represents the
South American Plate
Answer:
B
Explanation:
Where it shows on the map is B
A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,200 N. Find the speed with which a wave travels on this string.
Answer:
500 m/s
Explanation:
From the question,
V = √(T/m')................. Equation 1
Where V = Speed with which the wave travels on the string, T = Tension of the string, m' = Mass per unit length of the string.
Given: T = 1200 N, m' = 4.80×10⁻³ kg/m
Substitute these values into equation 1
V = √(1200/4.80×10⁻³)
V = √(250000)
V = 500 m/s
Hence the speed of the wave in the string is 500 m/s
a cat is being chased by a dog both are running in a straight line at constant speed. The cat has a headstart
Which correctly lists three places that fresh water is found?
rivers, oceans, lakes
oceans, coastal wetlands, seas
ice sheets, water vapor, rivers
water vapor, streams, oceans
The three places where fresh water can be found is ice sheets, water vapor, and rivers.
What is fresh water?A fresh water is any type of naturally occurring liquid or frozen water containing low concentrations of dissolved salts and other total dissolved solids.
Where fresh water can be foundFresh water can be found in the following locations;
ice sheets, water vapor, riversThus, the three places where fresh water can be found is ice sheets, water vapor, and rivers.
Learn more about fresh waters here: https://brainly.com/question/4432187
A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 27.5 s. (a) How tall is the tower
Answer: 187.68 m
Explanation:
Given
Time period [tex]T=27.5\ s[/tex]
The time period of a pendulum is given by
[tex]\Rightarrow T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
where L=height of tower
Put values
[tex]\Rightarrow 27.5=2\pi \sqrt{\dfrac{L}{9.8}}\\\\\Rightarrow L=\dfrac{27.5^2\times 9.8}{(2\pi )^2}=187.68\ m[/tex]
When a helicopter flies, what
type of energy is it using?
A. Thermal energy
B. Mechanical energy
C. Electromagnetic energy
Answer: B. Mechanical energy
Explanation: "Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position)."
"Potential and kinetic energy of an object as it is stored and moved. Examples of MECHANICAL ENERGY- roller coasters, hydraulic lift, bow and arrow, and helicopters. Helicopters, cars, roller coasters, all moving objects that are being acted on by a force."
Incorrect options: A. Thermal energy and B. Electromagnetic energy. (explanation on both below)
"Thermal energy (also called heat energy) is produced when a rise in temperature causes atoms and molecules to move faster and collide with each other. The energy that comes from the temperature of the heated substance is called thermal energy."
"Electromagnetic radiation, in classical physics, the flow of energy at the universal speed of light through free space or through a material medium in the form of the electric and magnetic fields that make up electromagnetic waves such as radio waves, visible light, and gamma rays."
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, calculate the time taken for the echo of the sound to return to the bat.
Answer:
0.15 s
Explanation:
From the question given above, the following data were obtained:
Speed of sound (v) = 330 m/s
Distance (x) = 25 m
Time (t) =?
The time taken for the echo of the sound to the bat can be obtained as follow:
v = 2x / t
330 = 2 × 25 / t
330 = 50 / t
Cross multiply
330 × t = 50
Divide both side by 330
t = 50 / 330
t = 0.15 s
Thus, it will take 0.15 s for the echo of the sound to the bat
The acceleration of a moving object is equal to
Answer:
Acceleration = Δv/Δt or change in velocity over change in time
Explanation:
What holds the moon in orbit around the earth?
A. The sun's gravity
B. The Earth's gravity
Answer:
Earth's gravity
Explanation:
hope this helps
Answer:
the earth gravity
Explanation:
Gravitational attraction provides the centripetal force needed to keep planets in orbit around the Sun and all types of satellite in orbit around the Earth. The Earth's gravity keeps the Moon orbiting us.
Part A
Playing in the street, a child accidentally tosses a ball (mass m) with a speed of v=18 m/s toward the front of a car (mass M) that is moving directly toward him with a speed of V=20 m/s . Treat this collision as a 1-dimensional elastic collision. After the collision, the ball is moving with speed v′ back toward the child and the car is moving with speed V′ in its original direction.
Part B
When we combine the equation from Part A with the conservation of momentum equation, we can solve for both final speeds. This relationship will involve the masses of the ball and the car, but we can apply a simplifying assumption: the car is so massive compared with the ball that its speed will not change at all as a result of this collision. Translate this sentence into an equation, what is V′ equal to? Now, having made this assumption, it becomes possible to solve the equation from Part A for the final speed of the ball, what is it?
Answer:
v' = -18 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved, as follows:[tex]p_{o} = p_{f} (1)[/tex]
The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):[tex]m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)[/tex]
The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):[tex]-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)[/tex]
If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).So, we can write again (3) as follows:[tex]-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)[/tex]
Replacing (2) and (4) in (1), we get:[tex]m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)[/tex]
Simplifying, and rearranging, we can solve for v'b, as follows:[tex]v'_{b} = -18 m/s (6)[/tex], which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had inverted its speed after the collision.to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on the end in a direction perpendicular to the wall until the block does not slip downward. If the spring is compressed, what is the coefficient of static friction
Answer:
μ = mg/kx
Explanation:
Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.
Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.
N = F' = kx
So, F = μN = μkx
μkx = mg
So, μ = mg/kx
two identical springs, each with a spring force constant k, are attached end to end. If a weight is hung from a single spring, it stretches the spring by a distance d. When this same mass is hung from the end of the two springs, which, again, are connected end-to-end, the total stretch of these springs is
Answer:
Δx = 2*d
Explanation:
According to Hooke's Law, in order to the mass be in equilibrium, when attached to one spring, no net force must act on it, so the algebraic sum of the elastic force and gravity must be zero, as follows:[tex]k*\Delta x = m*g (1)[/tex]
If we hang the mass from the end of the two springs attached end to end, in order to be in equilibrium, the total elastic force must be equal to gravity, as we have already said.We can express this elastic force, as the product of a Keff times the distance stretched by the two springs combined, as follows:[tex]F = k_{eff} * \Delta x_{eff} = m*g (2)[/tex]
Due to F is a tension, it will be the same at any point of the chain of springs, so we can write the following expression, for the distance stretched by any of the springs:[tex]\Delta x_{1} = \frac{F}{\ k_{1} } (3)[/tex]
The total distance stretched will be the sum of the distances stretched by any spring individually:[tex]\Delta x_{} = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (4)[/tex]
Replacing (4) in (2) and rearranging, we have:[tex]\frac{F}{k_{eff} } = \frac{F}{\ k_{1} } + \frac{F}{\ k_{2} } (5)[/tex]
Since k₁ = k₂ = k, we can find keff, as follows:
[tex]k_{eff} = \frac{k^{2} }{2*k} = \frac{k}{2} (6)[/tex]
Replacing (6) in (2), and making (2) equal to (1) we finally get:[tex]F = \frac{k}{2} * \Delta x_{eff} = k*\Delta x = m*g (7)[/tex]
Solving for Δxeff:[tex]\Delta x_{eff} = \frac{2*k*\Delta x}{k} = 2* \Delta x = 2*d (8)[/tex]
what is the tension on the rope
Answer:
Tor mata kobic furi♀️♂️❤
What do we call the part of the wave labeled with the letter
D in the image below?
B
MONO
A. Compression
B. Rarefaction
D. Wave speed
C. Frequency
Answer:
C. Frequency..............
A land breeze forms when:
a) cool air moves down off a mountain towards the coast
b) warm air rises up a mountain slope
c) cool air moves onto a shore from a large body of water
d) warm air from the land moves towards the water
Explanation:
Its D. The warm air from the land moves towards the water
Calculate the size of the magnetic field 20 m below a high voltage power line. The line carries 450 MW at a voltage of 300,000 V. Group of answer choices 0.237 T 0.0237 T 0.474 T 2.37 T 0.237 J
Answer:
[tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].
Explanation:
Power carried by the line [tex]=P=450 \mathrm{MW}=450 \times 10^{6} \mathrm{~W}[/tex]
Voltage across the line Volts
Current flowing in the line =i
Size of magnetic field =B
Distance from the line
Formula Used:
Current flowing is given as
[tex]i=\frac{P}{\Delta V}[/tex]
Magnetic field by the current carrying wire is given as
[tex]B=\left(\frac{\mu}{4 \pi}\right)\left(\frac{2 i}{r}\right)[/tex]
Inserting the values
[tex]B=\left(10^{-7}\right)\left(\frac{2(1500)}{(20)}\right) \\ B=1.5 \times 10^{-5} \mathrm{~T}[/tex]
Conclusion:
Thus, the magnetic field comes out to be [tex]1.5 \times 10^{-5} \mathrm{~T}[/tex].
Based on the image which parachuter will fall fastest
A
B
Or C
Explanation:
c willl fall fast then a and b
A proton moves at a speed of 0.12 x 10^7 m/s
at right angles to a magnetic field with a
magnitude of 0.58 T.
Find the magnitude of the acceleration of
the proton. The elemental charge is 1.60 x
10^-19 C.
Answer in units of m/s?.
Answer:
13/23 ydfffgggggggffttf
What happens during heat
transfer?
A. Heat always flows from cool to warm.
B. Heat always flows from warm to cool.
C. Heat always flows from warm to hot.
Answer:
B
Explanation:
According to Newton's 3rd Law of Motion, Doug, a baseball
player hits a ball with his bat with a force of 1,000N. The ball
exerts a reaction force equally against the bat of
A.less than 1,000N
B.more than 1,00N
C.1,000N
D.double 1,000N
Why does ice reflect more energy compared to water?
Answer:
while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows
Explanation:
so ice reflect more energy compared to water
A thin uniform-density rod whose mass is 3.0 kg and whose length is 2.6 m rotates around an axis perpendicular to the rod, with angular speed 37 radians/s. Its center moves with a speed of 13 m/s. (a) What is its rotational kinetic energy
Answer:
rotational kinetic energy is 1156.81 J
Explanation:
Given the data in the question;
first we find the moment of inertia of the rod as axis passes through the middle or center;
[tex]I = \frac{1}{12}ml^2[/tex]
where m is the mass of the road( 3.0 kg)
[tex]l[/tex] is the length of the rod ( 2.6 m )
so we substitute
[tex]I = \frac{1}{12}[/tex] × 3 × (2.6)²
[tex]I[/tex] = 1.69 kg.m²
now, to calculate the rotational kinetic energy,
kinetic energy due to rotation is;
KE[tex]_R[/tex] = [tex]\frac{1}{2}Iw^2[/tex]
where [tex]I[/tex] is moment of inertia( 1.69 kg.m² ),
w is angular frequency ( 37 radians/s )
we substitute
KE[tex]_R[/tex] = [tex]\frac{1}{2}[/tex] × 1.69 × ( 37 radians/s )²
KE[tex]_R[/tex] = 1156.81 J
Therefore, rotational kinetic energy is 1156.81 J
The rotational kinetic energy will be 1156.81 J. It is half of the product of the moment of inertia and angular speed.
What is rotational kinetic energy?The rotational kinetic energy is the kinetic energy generated by an object's rotation and It is a component of its overall kinetic energy.
The given data in the problem is;
m is the mass = 3.0 kg
l is the length = 2.6 m
[tex]\rm \omega[/tex] is the angular speed = 37 radians/s
v is the speed = 13 m/s
[tex]\rm E_k[/tex] is the rotational kinetic energy=?
The momentum of inertia is found by;
[tex]\rm I = \frac{1}{12} ml^2 \\\\ \rm I = \frac{1}{12} \times 3 \times (2.6)^2 \\\\\ \rm I = 1.69 \kg.m^2[/tex]
The rotational kinetic energy is found by;
[tex]\rm E_k= \frac{1}{2}I (\omega)^2 \\\\ \rm E_k= \frac{1}{2}\times 1.69 (37)^2 \\\\ \rm E_k=1156.81 \ J[/tex]
Hence the rotational kinetic energy will be 1156.81 J.
To learn more about the rotational kinetic energy refer to the link;
https://brainly.com/question/19305456
A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that has a height of 8.0 m. When the car reaches the top of the second hill, its speed is 11 m/s. Determine the work done by non-conservative forces on the car as it travels from the top of the first hill to the top of the second hill.
Answer:
The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.
Explanation:
By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.
From top of the first hill to the bottom
[tex]m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}[/tex] (1)
From the bottom to the top of the second hill
[tex]\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}[/tex] (2)
Where:
[tex]m[/tex] - Mass of the roller coaster car, in kilograms.
[tex]v_{1}[/tex] - Speed of the roller coaster car at the bottom between the two hills, in meters per second.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h_{1}[/tex] - Height of the first top of the hill with respect to the bottom, in meters.
[tex]W_{1, loss}[/tex] - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.
[tex]v_{2}[/tex] - Speed of the roller coaster car at the top of the second hill, in meters per seconds.
[tex]h_{2}[/tex] - Height of the second top of the hill with respect to the bottom, in meters.
[tex]W_{2, loss}[/tex] - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.
By using (1) and (2), we reduce the system of equation into a sole expression:
[tex]m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}[/tex] (3)
Where [tex]W_{loss}[/tex] is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.
If we know that [tex]m = 175\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h_{1} = 18\,m[/tex], [tex]h_{2} = 8\,m[/tex] and [tex]v_{2} = 11\,\frac{m}{s}[/tex], then the work done by non-conservative force is:
[tex]W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right][/tex]
[tex]W_{loss} = 6574.75\,J[/tex]
The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.
4. While cleaning your bedroom, you move your mattress to vacuum underneath your bed. You use a force of 48 N to move the mattress 1.5 meters out of the way. How much work was done?
Answer:
72 J
Explanation:
Use the Work formula
W= F x d
Given:
F - 48 N
d - 1.5 m
Solution:
W= F x d
W= 48 N x 1.5 m
W= 72 J
Explain why Large change in temperature makes heat flow fast and a small change does not.
Answer: Because of the amount of energy
Explanation: Heat is an energy, and small amounts of it does not effect large areas, whereas a very large blast of heat energy can greatly effect the air
identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.
Answer:
1) Linear motion
2) Rotational motion
3) Rotational Motion
4) Random Motion ( The ball can be rolling in any direction)
5) Linear motion
The types of motion in each activity include the following:
Walking a long a hallway- Linear motion Motion of the blades of the fan- Rotational motionEarths rotation- Rotational motion Ball moving on the ground- Random motion Soldiers marching- Linear motion.What is Motion?
This involves an object or a body changing position over time. There are
different types of motion with different examples as can be seen above in
this scenario.
Read more about Motion here https://brainly.com/question/453639
what is the reaction of zinc nitrate mixed with silver. It's a lab and we are supposed to show what the difference, or any reaction including physical, that would happen.
Answer:
b
Explanation:
Equation for Specific latent heat
2. Bank robbers have pushed a 1000 kg safe to a second story floor-to-ceiling window. They plan to break the window, then lower the safe 3.0 m to their truck. They stack up 500 kg of furniture, tie a rope between the rope and the furniture, and then place the rope over a pulley. Then they push the safe out of the window. What is the safe speed when it hits the truck
Answer:
5.4 m/s
Explanation:
Given that
Mass of the safe, m1 = 1000 kg
Distance to lower the safe, d = 3 m
Mass of furniture, m2 = 500 kg.
Speed of the safe, v = ?
To get the final speed by the time that the safe hits the truck, we first find its acceleration.
The total mass of the system is M = 1000 + 500 kg = 1500 kg
One of the forces acting on the system is that of gravity, and it acts on the safe friction acting on the furniture. Using the formula, we have
= m1*g - mu*m2g
= 1000 * 9.81 - 0.5 * 500 * 9.81
= 7357.5 N
From this calculated weight, we find the acceleration.
Acceleration, a = F/m
Acceleration, a = 7357.5 / 1500
Acceleration, a = 4.905 m/s²
From the question, we know that the Initial speed = 0 m/s
So, employing the use of one of the equations of motion, we have
v² - u² = 2aS
v² - 0 = 2 * 4.905 * 3
v² = 29.43
v = √29.43
v = 5.4 m/s
Determine experimentally which rotational axis yields the maximum rotational inertia (i.e., moment of inertia) and which yields the minimum rotational inertia for the broom stick. Draw a picture of the broom stick with its axis of rotation for (i) the minimum rotational inertia and (ii) the maximum rotational inertia.
Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ = [tex]\frac{1}{12}[/tex] m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ = [tex]\frac{1}{2}[/tex] m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass