Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
What is the correct organization of living things, from smallest to largest?
Cells - Tissues - Organs - Organ Systems - Organism
Organs - Tissues - Cells - Organ Systems - Organism
Cells - Organs - Tissues - Organism - Organ Systems
Cells - Organism - Tissues - Organ Systems - Organs
Which is the best explanation for why Toms technique works ?
Chemical messengers that stimulate a specific cellular response.
Glucose
Hormones
Mitochondria
Nerves
Answer:
Explanation:
hormones. please mark me brainliest
Learning task 2: Using the information you gathered from Learning Task 1, make a concept web of the contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
A. Andre- Marie Ampere
B. Michael Faraday
C. Heinrich Herts
D. James Clerk Maxwell
E. Hans Christian Oersted
Answer:
The contributions of the following scientist in the DEVELOPMENT OF MAGNETIC THEORY
James Clerk Maxwell Hans Christian OerstedExplanation:
George Green was the first personality to formulate a mathematical principle of magnetism and electricity and his system created the framework for the work of different scientists such as William Thomson, James Clerk Maxwell, and others. Magnetism is the power exercised by magnets when they drag or deflect each other. Magnetism is produced by the movement of electric charges.
The contributions of James Clerk Maxwell and Hans Christian Oersted, et al in the DEVELOPMENT OF MAGNETIC THEORY are as follows:
They discovered that the speed at which electromagnetic waves traveled was similar to that of lightThey proved that there was a proportional connection between electricity and magnetismAccording to the given question, we are asked to show the contributions which the aforementioned scientists had in the development of the magnetic theory.
As a result of this, we can see that James Maxwell first developed this theory in the nineteenth century and the theory was modified by other scientists who made the framework for the electrical system and magnetism.
Read more here:
https://brainly.com/question/17913237
This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
(i) Including the water in its internal cavity, the squid has a total mass of 6.50 kg.
(ii) The mass of the water in its cavity is 1.75 kg.
(iii) In order to escape its predators, the squid needs to achieve an escape speed of 2.5 m/s.
Answer:
6.79 m/s
Explanation:
By applying the principle of conservation of momentum.
The total momentum = MV - mv = 0 (since the squid is beginning at rest)
the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg
= 4.75 kg
speed of the squid (V) = 2.5 m/s
mass of the water expelled (m) = 1.75 kg
speed of the water (v) = ???
∴
4.75 × 2.5 = 1.75 × v
[tex]v = \dfrac{4.75 \times 2.5}{1.75 }[/tex]
v = 6.79 m/s
If the force of gravity suddenly stopped acting on planets, they would
A.) spiral slowly towards the sun
B.) continue to orbit the sun
C.) move in straight lines tangent to thier orbits
D.) spiral slowly away from the sun
E.) fly straight away from the sun
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-quency will the car and passenger vibrate onthe springs? Answer in units of Hz. The acceleration of gravity is 9.81 m/s^2.
Answer:
The required frequency = 0.442 Hz
Explanation:
Frequency [tex]f = ( \dfrac{1}{2 \pi}) \omega[/tex]
where;
[tex]\omega = \sqrt{\dfrac{k}{m} }[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )[/tex]
However;
[tex]k = \dfrac{F}{x}[/tex] and;
mass [tex]m = m_{car } + m_{person}[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )[/tex]
where;
[tex]F = m_{person}g[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )[/tex]
replacing the values;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )[/tex]
[tex]\mathbf{f = 0.442 \ Hz}[/tex]
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substance by 1 °C. Specific heat capacity can be calculated using the following equation:
q = mc deltaT
In the equation q represents the amount of heat energy gained or lost in joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and AT is the temperature change of the substance in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
Solve: When you mix two substances, the heat gained by one substance is equal to the heat lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with 1,000 g of water. The water changes temperature by 2 °C and the aluminum changes temperature by -74.95 °C.
A. Water has a known specific heat capacity of 4.184 J/g °C. Use the specific heat equation to find out how much heat energy the water gained (q).
B. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. Aluminum's accepted specific heat value is 0.900 J/g °C. Use this value to check your work.
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
The heat gained by the water can be obtained solving the following equation:[tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]
where cw = specific heat of water = 4.184 J/gºCm= mass of water = 1,000 gΔT = 2ºC Replacing these values in (1) we get:[tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]
B)
Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:[tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]
⇒ [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]
[tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]
which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.
what is the result of seafloor spreading?
Answer:
Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.
Explanation:
eventually the crust cracks.
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
PLEASEEEEEEEEE HELPPPPPPPPPP
Describe what determines magnetism.
Answer:
Magnetism is caused by the motion of electric charges. Every substance is made up of tiny units called atoms. Each atom has electrons, particles that carry electric charges. ... Their movement generates an electric current and causes each electron to act like a microscopic magnet.
Explanation:
Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
If California experienced heavy rainfall, which system would be responsible for it and WHY?
Answer:
California has one of the most variable climates of any U.S state, and often experiences very wet years followed by extremely dry ones . The state's reservoirs have insufficient capacity to balance the water supply between wet and dry years.
PLEASE MARK ME AS BRAINLIEST
A hot-air balloon has a volume of 2500 m3 . The balloon fabric (the envelope) weighs 860 N . The basket with gear and full propane tanks weighs 1200 N .
a) If the balloon can barely lift an additional 3400 N of passengers, breakfast, and champagne when the outside air density is 1.23kg/m3, what is the average density of the heated gases in the envelope?
Answer:
1.007 kg/m³
Explanation:
Given that:
volume = 2500 m³
density of air [tex]\rho = 1.23 \ kg/m^3[/tex]
weight of air displaced = [tex]V \times \rho \times g[/tex]
= 2500 × 1.23 × 9.81
= 30165.75 N
weight of the ballon fabric = 860 N; &
The propane weight = 1200 N
The passenger additional weight = 3400 N
weight of the heated gas will be = V × d × g
= 2500 × d 9.81
= 24525 d
For floating
The weight of air displaced is less than 30165.75
∴ 860 + 1200 + 3400 + 24525 d = 30165.75
5460 + 24525 d = 30165.75
24525 d = 24705.75
d = 24705.75 / 24525
d = 1.007 kg/m³
Hence, the average density of the heated gas = 1.007 kg/m³
What is nature/nurture debate and why is it important in psychology
A certain force gives object m1 an acceleration of 12.0 m/s2. The same force gives object m2 an acceleration of 3.30 m/s2. What acceleration would the force give to an object whose mass is (a) the difference between m1 and m2 and (b) the sum ofm1 andm2
Answer:
a) a = 4,552 m / s², b) a = 2,588 m / s²
Explanation:
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
a) acceleration
m = m₂-m₁
we substitute
a = [tex]\frac{F}{m_2 - m_1}[/tex]
1 / a = [tex]\frac{m_2}{F} - \frac{m_1}{F}[/tex]
let's calculate
[tex]\frac{1}{a}[/tex] = [tex]\frac{1}{3.3} - \frac{1}{12}[/tex]
[tex]\frac{1}{a}[/tex] = 0.21969
a = 4,552 m / s²
b) m = m₂ + m₁
a = F / (m₂ + m₁)
[tex]\frac{1}{a} = \frac{m_2}{F} + \frac{m_1}{F}[/tex]
we substitute
[tex]\frac{1}{a} = \frac{1}{3.3} + \frac{1}{12}[/tex]
a = 2,588 m / s²
A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
A weightlifter curls a 32 kg bar, raising it each time a distance of 0.50 m. How many times must he repeat this exercise to burn off the energy in one slice of pizza? Assume 25% efficiency.
Answer:
Explanation:
Average energy contained by a slice of pizza is 860 J .
energy used in lifting 32 kg bar by .50 m = mgh
= 32 x 9.8 x .5 = 156.8 J
efficiency is 25 % , so energy used up = .75 x 156.8 = 117.6 J
So number of times exercise to be repeated to burn off energy of a slice of pizza
= 860 / 117.6
= 7.3 or 7 times .
PLZ HELP WILL MARK BRAINLIEST!!
Amy has a mass of 50 kg, and she is riding a skateboard traveling 10 meters per second. What is her momentum?
5 kg·m/s
10 kg·m/s
50 kg·m/s
500 kg·m/s
Answer:
[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]
Explanation:
The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].
Answer:
500
Explanation:
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 32.7 m/s2 with a beam of length 5.29 m , what rotation frequency is required
Answer:
The rotation frequency required is 23.78 RPM
Explanation:
Given;
radial acceleration, a = 32.7 m/s²
length of the beam, r = 5.29 m
The linear velocity is calculated as;
[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar}[/tex]
where;
v is linear velocity
The angular velocity is calculated as;
[tex]\omega = \frac{v}{r} \\\\Recall, v = \sqrt{ar} \\\\Then, \omega = \frac{\sqrt{ar}}{r}} \\\\ \omega = \frac{\sqrt{32.7 \times5.29}}{5.29}\\\\\omega = 2.49 \ rad/s\\\\Angular \ frequency \ is \ calculated \ as;\\\\\omega = 2\pi f\\\\f = \frac{\omega}{2\pi} \\\\f = \frac{2.49}{2\pi} \\\\f = 0.396 \ rev/s\\\\f = 23.78 \ rev/min[/tex]
Therefore, the rotation frequency required is 23.78 RPM
True or False. Facts are based on observations. *
True
False
Answer:
TRUE
Explanation:
IM SMART
Answer:
Stay Safe! ,God bless you . The answer is false ,
Explanation:
The graph shows projected changes in the populations of the world.
World Population Growth
20,000
10,000
World
5,000
Asia
2,000
Africa
1,000
Europe
500
200
United States,
Canada, and Greenland
100 Mexico
Central America, Caribbean Islands,
50
and South America
Oceania (Australia and
20
nearby islands in the Pacific)
10
2040
2050
Based on the information in the graph, which region is expected to have the
greatest increase in its population over the period shown?
1950
1960
1970
1980
1990
2000
2010
2020
2030
Answer: C. Africa
Explanation:
The data given on the graph shows that:
Asia will grow from around 1,300 million to 5,000 million in 2050 which is an increase of:
= 5,000 /1,300 = 3.84 times
Europe will decrease over that period.
Africa will go from around 300 million to 2,000 million which is an increase of:
= 2,000 / 300
= 6.67 times
Mexico
, Central America, Caribbean Islands:
= 900 / 120
= 7.5 times
United States, Canada, and Greenland:
= 400/120
= 3.33
Oceania:
= 50 / 13
= 3.85
From the options give, Africa will see the greatest increase at 6.67 times its population in 1950.
Water will expand more than ___
A. door
B. juice
C. air
Answer:
Its A. door because liquids expand better than solids
3. A car is traveling up a 3% grade, with the speed of 85mph, on a road that has good, wet pavement. A deer jumps out onto the road and the driver applies the brakes 290-ft from it. The driver hits the deer at a speed of 20mph.If the driver did not have antilock brakes, and the wheels were locked the entire distance, would a deer-impact speed of 20mph be possible
Answer:
Explanation:
From the given information;
Let assume that:
the wheel radius = 15 inches
the driveline slippage = 3%; &
the gear reduction ratio (overall) = 2.5 to 1
So; using the equation:
[tex]v_1= \dfrac{2 \pi r n_o (1 -i)}{\varepsilon_o}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (1 -0.03)}{2.5}[/tex]
[tex]v_1= \dfrac{2 \times 3.14 \times \dfrac{15}{12} \times \dfrac{85}{100} (0.97)}{2.5}[/tex]
[tex]v_1 = 126.92 \ fp^3[/tex]
[tex]frl = 0.01 ( 1+ \dfrac{v}{147}) \ if \ v \ is \ ft/sec[/tex]
[tex]frl = 0.01 \Bigg( 1+ \dfrac{\dfrac{126.92 +(20)1.47 }{2} }{147}\Bigg)[/tex]
[tex]S = \dfrac{v_b ( v_1^r-v_2^r)}{2g(n_b \mu + frl \pm sin \ y}[/tex]
where;
[tex]\mu = 0.6[/tex]
[tex]291 = \dfrac{1.64( 126.92^2-29.9^2)}{64.4(n_b \times 0.6 +0.01532 +0.03}[/tex]
[tex]n_b = 1.33 \to which \ is \ not \ possible[/tex]
However;
[tex]n_b \mu = 1.33(0.6) = 0.80[/tex]
[tex]\mu = 0.9 \to[/tex] if the car's anti-clock breaking system did not fail
Thus;
[tex]n_b (0.9) = 0.80[/tex]
[tex]n_b =\dfrac{ 0.80}{(0.9) }[/tex]
[tex]n_b = 0.89[/tex]
Hence, the distance is possible if the anti-clock breaking system did not fail.
If 478 watts of power are used in 14 seconds,how much work was done
Answer:
6692J
Explanation:
Power is defined as the rate at which work is being done.
So,
Power = [tex]\frac{workdone}{time }[/tex]
Work done = Power x time
Given parameters:
Power = 478watts
Time = 14s
So;
Work done = 478 x 14 = 6692J
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
at what speed does the kg ball move ?
Answer: Choice A) 2 meters per second
=======================================================
Explanation:
The smaller ball has momentum of
p = m*v
p = (1 kg)*(4 m/s)
p = 4 kg*m/s
All of this momentum transfers into the larger ball because the smaller ball comes to a complete stop.
For the larger ball, we have p = 4 and m = 2. Let's find v.
p = m*v
4 = 2*v
4/2 = v
2 = v
v = 2 m/s which is why the answer is choice A
The larger ball moves at a speed of 2 meters per second. The speed is cut in half compared to the smaller ball because the larger ball has more inertia (aka more mass), and therefore it takes more energy to move it. If you apply the same energy to each, then the smaller object moves faster.
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
How do sound waves travel?