Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
What do you diagram to analyze orbital motion ?
The diagram to analyze the orbital motion can be shown as,
Here, a is the acceleration of moon and v is the speed.
The above diagram indicates the orbital motion of the moon around the earth. The moon is more towards the earth than the sun due to larger gravity of earth and at the same time the moon has its velocity that tends moon to move. Therefore, the moon has balanced gravitational and centripetal force to keep in an uniform orbital motion.
Examine the heating curve for water below. Answer each question andcomplete the table to review your understanding of heating curves
Explanation:
When the temperature is increasing, the kinetic energy change, and when the graph is horizontal (the water is changing phases), the potential energy changes.
So at A, C, and E, there is a change in kinetic energy and at B and D there is a change in potential energy.
Then, we use a specific heat capacity of 4.2 J/g°C when the water is at A, C, and E. We use latent heat of 334 J/g for melting and latent heat of 2260 J/g for evaporation.
Answer:
Therefore, the complete table is
a) your one friend and their bumper boat (m = 210 kg) are traveling to the left at 3.5 m/s and your other friend and their bumper boat (m = 221 kg) are traveling in the opposite direction at 1.8 m/s. The two boats then collide in a perfectly elastic one-dimensional collision. How fast is your first friend and their boat moving after the collision?b) after the collision, a constant drag force between the water and the boat causes your first friends boat to come to a stop. If the boat travels 2.7 m before stopping, what is the magnitude of the constant drag force?
ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]What is the shortest distance in which you can stop, after the brakes are applied, without the groceries sliding off the seat? The static and kinetic coefficients of friction are, respectively, 0.65 and 0.45. Assume that the surface of the seat is horizontal.
The shortest distance in which you can stop, after the breaks are applied is 80.38 m.
What is the shortest distance you can stop?The shortest distance in which you can stop is calculated by applying the principle of conservation of energy and work energy principle.
K.E = ¹/₂mv²
where;
K.E is your kinetic energy
m is your mass
v is your speed
The work done by force of friction before you stop is calculated as follows;
W = Ffx
where;
Ff is the frictional forcex is the shortest distance you can stopW = (μmg)x
where;
μ is coefficient of kinetic frictionW = K.E
(μmg)x = ¹/₂mv²
(μg)x = ¹/₂v²
x = (v²) / (2μg)
x = (32²) / (2 x 0.65 x 9.8)
x = 80.38 m
Thus, the shortest distance in which you can stop, after the breaks are applied is determined by applying the principle of conservation of energy.
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The complete question is below:
Your are driving at 32 m/s, what is the shortest distance in which you can stop, after the brakes are applied, without the groceries sliding off the seat? The static and kinetic coefficients of friction are, respectively, 0.65 and 0.45. Assume that the surface of the seat is horizontal.
What are the units for buoyant force?
Answer Unit of Buoyant Force The unit of the buoyant force is the Newton (N). = Here, F= buoyant force
Explanation:
i hope this helpes
3 Fig. 2.1 is a head-on view of an airliner flying at constant speed in a circular horizontal path. The centre of the circle is to the left of the diagram. Fig. 2.1 II (a) On Fig. 2.1, draw the resultant force acting on the airliner. Explain your answer.
Answer:
Part (a)
The airliner is moving at a constant speed in a circular path, so it follows a circular motion where the net force is the centripetal force that points to the center of the circle. So, we can draw the resultant force as
Part (b)
The weight and aerodynamic lift force is represented in the following free body diagram
Then, we can calculate the net vertical force and the net horizontal force.
The net vertical force is
Fnety = (1.39 x 10⁶ N)cos(30) - (1.20 x 10⁶ N)
Fnety = 1.20 x 10⁶ N - 1.20 x 10⁶ N
Fnety = 0
The net horizontal force
Fnetx = (1.39 x 10⁶ N)sin(30)
Fnetx = 6.95 x 10⁵ N to the left.
Therefore, the net force has a component only in the horizontal direction which is equivalent to the direction shown in part (a)
Part (c)
By the second law of Newton, if there is a net force, there is acceleration. So, there is a change in the velocity. In this case, the speed is constant but the velocity is changing constantly because the airliner is changing its direction to follow the circular path.
A string with both ends fixed is vibrating in its second harmonic. The waves have a speed of 36m/s and a frequency of 60Hz. The amplitude of the standing wave at an antinode is 0.6cm. calculate the amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm on the left hand end of the string.
The amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm is 0.0547m, 0.027m and 0.0137m respectively.
Amplitude is the maximum range of vibration or oscillation, measured from the equilibrium position
According to the equation of the second harmonic motion
A = sin (kx)
A = Amplitude
k = [tex]\frac{2\pi f}{v}[/tex] = [tex]\frac{2*\pi * 60}{36}[/tex] = 10.467
x = distance of the point
For x = 30 cm = 0.3 m
A = sin (kx)
A = Sin (10.467 * 0.3)
A = 0.0547 m
For x = 15 cm = 0.15 m
A = sin (kx)
A = Sin (10.467 * 0.15)
A = 0.027 m
For x = 7.5 cm = 0.075 m
A = sin (kx)
A = Sin (10.467 * 0.075)
A = 0.0137 m
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A blink of an eye is a time interval of about 150ms for an average adult. The closure portion of the blink takes only about 55ms. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of 16.6 degree. What is the value of the angular acceleration the eyelid undergoes while closing 2. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is 1.25 cm?
ANSWER:
STEP-BY-STEP EXPLANATION:
The first thing is to convert the time into a second, just like this:
[tex]t=55\text{ ms}\cdot\frac{1\text{ s}}{1000\text{ ms}}=0.055\text{ s}[/tex]Now, convert the angular displacement of the eyelid from degrees to rad:
[tex]\partial\theta=16.6\text{\degree}\cdot\frac{2\pi\text{ rad}}{360\text{\degree}}=0.29\text{ rad}[/tex]We can calculate the angular velocity, dividing the angular momentum by the time, like this:
[tex]w=\frac{0.29}{0.055}=5.27\text{ rad/s}[/tex]The angular acceleration is calculated by means of the quotient of the difference in angular velocity and time, like this:
[tex]a_w=\frac{\delta w}{\delta t}=\frac{5.27-0}{0.15-0.055}=55.47\text{ rad/s}^2[/tex]the tangential acceleration would be:
Physics
Hi dears how could we solve this question basically how to plug it in calculator
The new length of the steel bar is 11.65 cm.
What is linear expansion?
Linear expansion can be defined as the increase in length of a material due to increase in the temperature of the material.
Mathematically, the linear expansion of a material is given as;
ΔL = L₀αΔT
where;
ΔL is the change in length or increase in lengthL₀ is the initial length of the steelα is the coefficient of linear expansion of steel = 11 x 10⁻⁶/⁰CΔT is change in temperatureThe change in the length of the steel is calculated as follows;
ΔL = (11.5 cm) x (11 x 10⁻⁶/⁰C) x (1221 ⁰C - 22 ⁰C)
ΔL = 0.152 cm
The new length of the steel bar is calculated as follows;
L = ΔL + L₀
L = (0.152 cm) + (11.5 cm)
L = 11.65 cm
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Which of the following water molecules have the greatest kinetic energy?Select one:a. Cool water.b. Warm water.c. Boiling water.d. They all have the same kinetic energy.
Since all the molecules in the boiling water will have more energy introduced by heat, then the molecules with the greatest kinetic energy are the ones from the boiling water.
As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 = Fp = F = (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 = Fp = F = (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 = r5 =
a) Recall, the net torque on the rod must be zero. Thus,
Σt = 0
where
t represents torque
Thus,
t1 + t2 - tr - t3 = 0
t = rF
where
F = force
r = distance
r1F1 + r2F2 - rrFr - r3F3 = 0
r3F3 = r1F1 + r2F2 - rrFr
r3 = (r1F1 + r2F2 - rrFr)/F3
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
Thus,
r3 = (r1m1g + r2m2g - rrmrg)/m3g
g cancels out
r3 = (r1m1 + r2m2 - rrmr)/m3
From the information given,
r1 = 10 cm = 10/100 = 0.1 m
r2 = 90 cm = 90/100 = 0.9 m
rr = 100/2 = 50 cm = 50/100 = 0.5 m
m1 = 281 g = 281/1000 = 0.281 kg
m2 = 177g = 0.177 kg
mr = 158g = 0.158 kg
m3 = 200g = 0.2kg
By substituting these values into the equation,
r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
r3 = 0.542 m
The force exerted by the pin, Fp = mg
g = 9.8
Fp = (m3 - mr - m1 - m2)g
Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8
Fp = - 0.981
Taking the absolute value,
IFpI = 0.981 N
F = - 90 degrees
b) r1F1 + r2F2 - rrFr - r4F4 = 0
r4F4 = r1F1 + r2F2 - rrFr = 0
F4 = (r1F1 + r2F2 - rrFr)/r4
Note,
F1 = T1 = m1g
F2 = T2 = m2g
F3 = T3 = m3g
F4 = T4 = m4g
Thus,
m4g = (r1m1g + r2m2g - rrmrg)/r4
m4g = (r1m1 + r2m2 - rrmr)/r4
r4 = 0.2
By substituting these values into the equation,
m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2
m4 = 0.542 kg
The force exerted by pin is
Fp = (m4 + mr - m1 - m2(g
Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8
Fp = 2.37 N
Fp = 2.37 N
F = 90 degrees
c) When the pin does not exert a force,
Fp = 0
F1 + F2 - Fr = F5
m1 + m2 - mr = m5
m5 = 0.281 + 0.177 - 0.158
m5 = 0.3 kg
Since the net torque on the rod is zero,
t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
t5 = t1 + t2 - tr - t5
r5 = r1F1 + r2F2 - ffFr)/F5
r5 = (r1m1 + r2m2 - rrmr)/m5
r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3
r5 = 0.36
You place a box weighing 276 N on an inclined plane that makes a 44.5° angle with the horizontal.
Compute the component of the gravitational force acting down the inclined plane.
Answer in units of N.
Answer:
this is the answer
Explanation:
hope it helps
What is the wavelength of the sound produced by a bat if the frequency of the sound is 90 kHz on a night when the air temperature is 22°C?
We know that the wavelength is related to the speed of the wave by:
[tex]v=\lambda f[/tex]where f is the frequency.
The speed of sound on air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]so in this case the speed is:
[tex]v=331\sqrt[]{1+\frac{22}{273}}=344.08[/tex]Plugging this and the frequency in the first expression above we have:
[tex]\begin{gathered} 344.08=90\times10^3\lambda \\ \lambda=\frac{344.08}{90\times10^3} \\ \lambda=3.82\times10^{-3} \end{gathered}[/tex]Therefore the wavelength is:
[tex]3.82\times10^{-3}\text{ m}[/tex]how many joules does a lamp marked 12 volts, 24 w consumed in an hour . And also what is the current?
Given data:
* The voltage given is 12 volts.
* The value of the power given is 24 Watt.
* The time till which the lamp is used is,
[tex]\begin{gathered} t=1\text{ hr} \\ t=60\times60\text{ s} \\ t=3600\text{ s} \end{gathered}[/tex]Solution:
(a). The energy consumed by the lamp in time t is,
[tex]E=P\times t[/tex]where P is the power of lamp, t is the time, and E is the energy consumed,
Substituting the known values,
[tex]\begin{gathered} E=24\times3600 \\ E=86400\text{ J} \\ E=86.4\times10^3\text{ J} \\ E=86.4\text{ kJ} \end{gathered}[/tex]Thus, the energy consumed by the lamp in 1 hour is 86.4 kJ.
(b). The power of the lamp in terms of voltage and current is,
[tex]\begin{gathered} P=VI \\ I=\frac{P}{V} \end{gathered}[/tex]where P is the power, V is the voltage and I is the current,
Substituting the known values,
[tex]\begin{gathered} I=\frac{24}{12} \\ I=2\text{ A} \end{gathered}[/tex]Thus, the current flowing through the lamp is 2 A.
What is the momentum of a 934 kg car moving 10 m/s?
ANSWER:
9340 kg*m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 934 kg
Speed (v) = 10 m/s
The formula to calculate the momentum is as follows:
[tex]\begin{gathered} p=m\cdot v \\ \text{ we replacing} \\ p=934\cdot10 \\ p=9340\text{ kg*m/s} \end{gathered}[/tex]The momentum of the car is 9340 kg*m/s.
The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley.
Answer: It should be the answer beginning like this
The linear spread is that
The radial acceleration of locations along the blade's outer edge is approximately 17580 [tex]m/s^2[/tex].
What is Radial acceleration?Radial acceleration describes the acceleration of an object travelling on a circular path towards the circle's center. It can be defined as the rate of change of tangential velocity with regard to time and is also known as centripetal acceleration.
Given:
A table saw's engine rotates at 3450 revolutions per minute.A V-belt connects a pulley that's attached to the motor shaft to a second pulley half the diameter.A 0.208 m circular saw blade is installed on the same rotating shaft as the second pulley.We know that the motor is rotating at 3450 rev/min. One revolution is equal to 2π radians, so we can convert the motor speed to radians per minute:
ω₁ = (3450 rev/min) x (2π rad/rev) = 21675π rad/min
The second pulley is half the diameter of the first pulley, so its angular speed, ω₂, is twice that of ω₁:
ω₂ = 2ω₁ = 43350π rad/min
The circular saw blade is mounted on the same shaft as the second pulley, so it also rotates at the same angular speed:
ω = ω₂ = 43350π rad/min
We can now calculate the linear speed of the small piece of wood moving at the same rate as the rim of the circular saw blade, indicated by v. The circumference of the circle is supplied by the rim of the circular saw blade:
C = πd = π(0.208 m) = 0.6548 m
The linear speed of the little piece of wood is equal to the tangential speed of the circular saw blade's rim:
v = ωr
where r is the circular saw blade's radius, given by half its diameter:
r = d/2 = 0.208/2 = 0.104 m
By substituting the values, we obtain:
v = r = (43350 rad/min) x (0.104 m) x (1/60) = approx. 23.0 m/s
As a result, the linear speed of the little piece of wood moving at the same rate as the rim of the circular saw blade is about 23.0 m/s.
Next, compute the radial acceleration of locations on the blade's outer edge, represented by. The radial acceleration is calculated as follows:
α = rω²
By substituting the values, we obtain:
r2 = (0.104 m) x (43350 rad/min)2 x (1/602) = 17580 m/s2 (approximate)
Therefore, the radial acceleration of points on the outer edge of the blade is approximately 17580 m/s². This high radial acceleration explains why sawdust doesn't stick to the teeth of the saw blade.
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Your question is incomplete, most probably the complete question is:
The motor of a table saw is rotating at 3450{\rm rev/min}. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208{\rm m}is mounted on the same rotating shaft as the second pulley.
The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed?
v =_______________________ m/s
Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
\alpha=______________________m/s2
Peter is trying to ignite the hotplate by turning the gas knob. Suppose thatthe minimum moment of couple about the center of the gas knob requiredto ignite the hotplate is 0.3 N m. Calculate the minimum force (F, and F2)that required to exert. Given that the diameter of gas knob is 5 cm.
the minimum force is 6 Newtons
Explanation
A moment of a force, or a torque, is a measure of a force's tendency to cause a body to rotate. The moment depends on both the force, and on the position at which the force acts, it is given by the expression
[tex]M=F\cdot d\text{ }[/tex]where F is the exerted force and d is the distance
Step 1
then, let
[tex]\begin{gathered} M=\text{0}.3\text{ Nm} \\ F=F \\ \text{distance}=\text{ 5 cm= }\frac{5}{100}m=0.05\text{ m} \\ \end{gathered}[/tex]now, replace in the formula
[tex]\begin{gathered} M=F\cdot d\text{ } \\ 0.3\text{ Nm=F}\cdot0.05\text{ m} \\ \text{divide both sides by 0.05 m} \\ \frac{0.3\text{ Nm}}{0.05\text{ m}}\text{=}\frac{\text{F}\cdot0.05\text{ m}}{0.05\text{ m}} \\ 6N=F \end{gathered}[/tex]therefore, the minimum force is 6 Newtons
I hope this helps you
If a bus is traveling at 12m/s and a passenger on the bus is walking to the back of the bus at a velocity of 5m/s, what is the relative velocity of the passenger relative to the ground?
The relative velocity is 17m/s.
The relative velocity of the passenger relative to the ground can be found by applying the concept of relative motion.
speed of bus (vb)=12m/s
speed of passenger inside the bus(vp)= 5m/s opposite to the speed of bus
speed of passenger relative to the ground = v
v= vb+vp
v= 12+(-5), since passenger is traveling in opposite direction
v=7m/
Therefore, the velocity of passenger relative to the ground is 7m/s.
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Which of the following is true for an isolated system? I. Matter is able to freely enter or exit the system.II. Heat is able to freely enter or exit the system.III. Work is able to freely enter or exit the system.II onlyNone of the aboveI or II onlyI only
Note that, an isolated system does not allow:
• the exchange of energy
,• the exchange of matter
Therefore, in an isolated system, neither
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I need help with this question There is 4 answersa)b)c)d)
Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
[tex]\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
[tex]\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
[tex]\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}[/tex]Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
[tex]\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Thus the total momentum of the system, in this case, is 143.5 kg m/s
some more disconnectiob
let the student report the tutor and check
answer is updated
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Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net. With what speed did he leave the floor?
Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net, the speed from which he would have left the floor would be 4.64 m / s .
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net.
By using the third equation of the motion,
v² - u² = 2 × a × s
0 - u² = 2 × -9.81 × 1.10
u = 4.64 m / s
Thus, the speed from which he would have left the floor would be 4.64 m / s .
To learn more about equations of motion here, refer to the link given below;
brainly.com/question/5955789
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Can you help me match these with the correct word
When an object is thrown up into the air the time up is equal to the time down.
When an object is thrown up into the air and reacher the apex, the velocity of the object is zero.
If you throw an object straight up the air the velocity of the object is decreasing on its way up.
A roller coaster car begins its roll from the top of the tracks at a speed of2 meters per second. When it reaches the bottom of the 200-meter drop four seconds later, its speed is 22 meters per second. What was the averagespeed of the roller coaster ride in meters per second over the 200-meter drop
The speed of the car from the top of the track is,
[tex]u=2ms^{-1}[/tex]The distance traveled by the car is,
[tex]d=200\text{ m}[/tex]The speed of the car after 4 seconds is,
[tex]v=22ms^{-1}[/tex]Thus, the average speed of the car is,
[tex]v_{av}=\frac{u+v}{2}[/tex][tex]\begin{gathered} v_{av}=\frac{2+22}{2} \\ v_{av}=12ms^{-1} \end{gathered}[/tex]Thus, the average speed of the roller coaster car is 12 meter per second.
Need help 82x2682 please
ANSWER:
STEP-BY-STEP EXPLANATION:
We have the following multiplication:
[tex]undefined[/tex]If a space ship traveling at a 1000 miles per hour enters and area free of Gravitational forces,it’s engine must run at some minimum level in order to maintain the ships velocity. is this statement true or false
The given statement is false.
If a spaceship traveling at 1,000 miles per hour enters an area free of gravitational forces, its engine must run at some maximum level in order to maintain the ship’s velocity
Which resistors in the circuit must always have the same current?A.B and CB.A and BC.C and DD.A and D
ANSWER
D. A and D
EXPLANATION
Two resistors have the same current if they are connected in series. As we can see in the schematic, resistors B and C are connected in parallel, so they don't have the same current - unless they have the same resistance.
Resistors A, D, and the equivalent of the parallel resistors (B and C) are connected in series, so they always have the same current.
Hence, of these options, resistors A and D always have the same current.
The total mechanical energy of the roller coaster cart below at Point A is 180,000 J. The speed of the cart at Point B is +20 m/s. Assume no energy is lost due to dissipative forces such as friction. A) What is the mass (in kg) of the roller coaster cart? B) What is the potential energy at Point A? C) What is the kinetic energy at Point A?
Mechanical energy (ME) = Potential energy(PE) + kinetic energy (KE)
PE = mgh
m= mass
g= gravity
h= height
KE= 1/2 m v^2
v= speed
Point B
ME = KE + PE
PE = 0 (height = 0 )
KE = 1/2 (m) v^2
180,000 = 1/2 (m) (20)^2
m = 180,000 / (1/2 (20)^2 )
m= 900 kg
Point A.
ME = 180,000 J = PEa + KE a
PEa = m g h = 900 (9.8) (20) = 176,400J
MEa = PEa + KEa
KEa = MEa - PEa = 180,000 - 176,400 J = 3,600 J
A) mass = 900 kg
B) 176,400 J
C) 3,600 J
calculate how much heat is released when 500g of platinum is cooled from 250.0K to 240.0K
Answer:
Q = 665 J
Explanation:
Given:
m = 500 g = 0.5 kg
T₁ = 250.0 K
T₂ = 240.0 K
c = 133 J / ( kg*⁰K) - Specific heat capacity of platinum
___________
Q - ?
Heat is released:
Q = c*m*( T₁ - T₂) = 133*0.5*(250.0 - 240.0) =
= 133*0.5*10 ≈ 665 J
What is the initial velocity of an automobile acquiring a final velocity of 32 m/s with an acceleration of 4.0m/s ²
Answer:
Explanation:
Given:
V = 32 m/s
a = 4.0 m/s²
__________
V₀ - ?
V = V₀ + a*t
V₀ = V - a*t = 32 - 4*t
Time is not set according to the condition of the problem!
There's not enough given information t o answer the question. It depends on how long the car has been accelerating.
it could be 28 m/s 1 second ago.
it could be 16 m/s 4 seconds ago.
it could be 10 m/s 5.5 seconds ago.
etc.
i'll take a wild guess and speculate that the question actually tells how long the car has been accelerating, but you didn't copy that part.
