Answer:
dy
Explanation:
My guess of guts.
okay bye
g A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04t^3−0.09t^2 Correct: Your answer is correct. (b) What is the velocity after 1 second(s)? v(1) = -0.05 Correct: Your answer is correct. ft/s (c) When is the particle at rest? t = 0 Correct: Your answer is correct
Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
[tex]v(t) =\frac{d(f(t))}{dt}[/tex]
[tex]v(t) = 4(0.01)t^{4-1} - 3(0.03)t^{3-1}\\v(t) = 0.04t^3 - 0.09t^2[/tex]
Hence the velocity of the particle at time t is [tex]v(t) = 0.04t^3 - 0.09t^2[/tex]
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
[tex]v(t) = 0.04t^3 - 0.09t^2\\v(1) = 0.04(1)^3 - 0.09(1)^2\\v(t) = 0.04 - 0.09\\v(t) = -0.05[/tex]
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second
Complete each statement by dragging the forms of energy into their appropriate boxes.
wind turbine
roller coaster going downhill
toaster
car
A
converts electrical energy into thermal energy.
A
converts rotational energy into electrical energy.
A
converts gravitational energy into mechanical energy.
A
converts rotational energy into mechanical energy.
Statements 1,2,3 and 4 match statements B, C, A, and D respectively.A wind turbine converts rotational energy into electrical energy.
What is the law of conservation of energy?According to the Law of conservation of energy. Energy can not be created nor be destroyed, it can transfer from one to another form.
1.A wind turbine converts rotational energy into electrical energy.
2.A roller coaster going downhill converts gravitational energy into mechanical energy
3. Toaster converts electrical energy into thermal energy
4.A car converts rotational energy into mechanical energy.
Hence,statements 1,2,3 and 4 match statements B, C, A, and D respectively.
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Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be
Answer:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:
[tex] L_{i} = L_{f} [/tex]
[tex] I_{i}\omega_{i} = I_{f}\omega_{f} [/tex] (1)
The initial angular speed is:
[tex] \omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d} [/tex]
The moment of inertia (I) of a sphere is:
[tex] I = \frac{2}{5}mr^{2} [/tex] (2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:
[tex] \frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f} [/tex]
[tex]9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}[/tex]
[tex]\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s[/tex]
Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!
Question C) needs to be answered, please help (physics)
(a) Differentiate the position vector to get the velocity vector:
r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k
v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j
(b) The velocity at t = 2.00 s is
v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j
(c) Compute the electron's position at t = 2.00 s:
r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k
The electron's distance from the origin at t = 2.00 is the magnitude of this vector:
||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that
tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º
or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.
what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec
Answer:
F = 58.35 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.
We have to convert speeds from kilometers per hour to meters per second
[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]
[tex]v_{f}=v_{o}+(a*t) \\[/tex]
where:
Vf = final velocity = 15 [m/s]
Vi = initial velocity = 7.22 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Note: the positive sign of the above equation is because the car increases its speed
15 = 7.22 + (a*4)
a = 1.945 [m/s^2]
Now we can use the Newton's second law:
F = m*a
F = 30*1.945
F = 58.35 [N]
3- For given three vectors a, b and c, c = a x b, then the vector c is:
Answer:
VB
Explanation:
Question 1-1: In each case, lifting or pushing, why must you exert a force to keep the object moving at a constant velocity?
Answer:
We must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
Explanation:
LIFTING:
When an object is lifted, we first need to overcome the force exerted on it by the field of gravity. Due to this force, which is also called the weight of object, we must apply a force on the object to keep it moving at constant speed, otherwise the gravity force will cause the object to slow down and eventually fall back on ground.
PUSHING:
When pushing an object the person must apply the force to first overcome the frictional force. The frictional force acts in opposite direction of motion. Thus, to move the object at constant speed we must apply force to it.
Hence, we must apply a force to keep the object moving at a constant velocity due to gravitational force or weight (in case of lifting), and due to frictional force (in case of pushing).
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
PLEASE HELP WILL MARK BRAINLIEST
Answer:
also choose D and E i belive those are also correct
It takes 525 J of work to compress a spring 25 cm. What is the force constant of the spring (in kN/m)?
Answer:
1.680kN/m
Explanation:
Work done by the spring is expressed as shown:
[tex]W = \frac{1}{2}ke^2[/tex] where:
k is the spring constant
e is the extension
Given
W = 525Joules
extension = 25cm = 0.25m
Substitute into the formula:
[tex]525 = \frac{1}{2}k(0.25)^{2} \\525 = \frac{0.0625k}{2}\\ 525 = 0.03125k\\k = \frac{525}{0.3125}\\k = 1680N/m\\k = 1.680kN/m[/tex]
Hence the force constant of the spring is 1.680kN/m
What are the four basics for knowledge called "innate knowledge"?
Answer: The four basics of innate knowledge are as follows:
Explanation:
1. This knowledge is learned from birth and it is inborn knowledge. It is allows the organism to act naturally. For example, a dog is not taught to pant, but it pants to reduce heat from the body.
2. It is inherent.
3. It is essential for survival.
4. It arises from intellectual knowledge rather than being learned via experiences.
A ball is thrown upward from the ground with an initial speed of 20.6 m/s; at the same instant, another ball is dropped from a building 14 m high. After how long will the balls be at the same height above the ground?
Answer:
t= 0.68 s
Explanation:
Neglecting air resistance, both balls are only under the influence of gravity, so we can use the kinematic equation for vertical displacement for both balls.First of all, we define two perpendicular axes, coincidently with horizontal and vertical directions, that we denote as x-axis and y-axis respectively.Assuming that the upward direction is the positive one, g must be negative as it always points downward.Taking the ground as our zero reference for the vertical axis (y axis), the equation for the ball thrown upward can be written as follows:[tex]y = v_{o}* t -\frac{1}{2} * g * t^{2} (1)[/tex]
As the second ball is dropped, its initial velocity is 0. Taking the height of the building as the initial vertical position (y₀), we can write the equation for the vertical displacement as follows:[tex]y = y_{o} - \frac{1}2}*g*t^{2} (2)[/tex]
As the left sides of (1) and (2) are equal each other (the height of both balls above the ground must be the same), the time must be the same also.We can rearrange (2) as follows:[tex]y -y_{o} = -\frac{1}{2}*g* t^{2} (3)[/tex]
Replacing the right side of (3) in (1), we get:[tex]y = v_{o}*t + (y- y_{o})[/tex]
⇒ [tex]t =\frac{y_{o} }{v_{o} } =\frac{14 m}{20.6 m/s} = 0.68 s[/tex]
⇒ t = 0.68s
why the bodies of water important for recreation
Explanation:
Recreational water activities can have substantial benefits to health and well-being. Swimming pools, beaches, lakes, rivers and spas provide environments for rest and relaxation, physical activity, exercise, pleasure and fun. Yet they also present risks to health.
pls make it the brainliest of it has helped you !!!!
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.1 m in a time of 1.8 s. The readout on the display indicates that the average power you are producing is 90 W. What is the magnitude of the force that you exert on the handle?
Answer:
147.27N
Explanation:
Power = workdone/time
Power = Force*distance/time
Given
Power = 90Watts
Distance = 1.1m
Time = 1.8secs
Force = ?
Substitute the given parameters into the formula:
[tex]90 = \frac{1.1d}{1.8}\\cross \ multiply\\ 90 \times 1.8 = 1.1F\\162 = 1.1F\\1.1F = 162\\F = \frac{162}{1.1} \\F = 147.27N[/tex]
Hence the magnitude of the force that you exert on the handle is 147.27N
Peter rides his bike 5 blocks north and then 10 blocks east. What is his displacement? What is his total distance traveled?
Answer:
5 blocks east. His total distance is 15 feet.
Explanation:
I hope that this helps! Have a good day!
We recommend that our students get at least _____ hours of behind-the-wheel instruction.
A. 6
B. 10
C. 25
D. 50
A 6.13-g bullet is moving horizontally with a velocity of 361 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1233 g, and its velocity is 0.741 m/s after the bullet passes through it. The mass of the second block is 1646 g. (a) What is the velocity of the second block after the bullet imbeds itself
Answer:
v₃ = 0.786 m/s
Explanation:
Here, we will use the law of conservation of momentum, which states the following:
Total Momentum of System Before Collision =
Total Momentum of System After Collision
m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃
where,
m₁ = mass of bullet = 6.13 g = 0.00613 kg
m₂ = mass of 1st block = 1233 g = 1.233 kg
m₃ = mass of 2nd block = 1646 g = 1.646 kg
u₁ = speed of first bullet before collision = 361 m/s
u₂ = speed of first block before collision = 0 m/s
u₃ = speed of 2nd block before collision = 0 m/s
v₁ = speed of bullet after collision
v₂ = speed of 1st block after collision = 0.741 m/s
v₃ = speed of 2nd block after collision = ?
Therefore,
(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)
1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)
since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)
Therefore,
1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)
v₃ = (1.2992 kg m/s)/(1.6521 kg)
v₃ = 0.786 m/s
You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?
1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.
Answer:
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
Explanation:
centripetal force = m ω² R
here m is mass , ω is angular velocity and R is distance of penny from centre
So this force depends upon R
penny on the outer edge will require greater centripetal force to move in circular path .
The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .
Omar observes that many buildings in his city were built using limestone. He has read that acid rain can damage limestone. He also knows that limestone reacts with acids and that chemical reactions are affected by temperature. With this in mind, he conducts an investigation to see how the amount of damage to the limestone is affected by the amount of acid on the stone.
Which statement describes the correct plan for his procedure?Rhonda watched a video taken by a camera that was lifted into the upper atmosphere by a weather balloon. She saw the balloon pop when it reached a certain height. Afterward, Rhonda wondered what effect the air pressure at high altitudes has on the volume of gas in balloons.
What scientific practice is Rhonda performing
Answer:he conducts the investigation to see the effect acidic water has on limestone
Explanation:
What is the probability that a junior non-Physics major and then a freshman non-Physics major are chosen at random?
Answer:
Probability = 0.0244
Explanation:
Probability that Junior Non Physics Major & then a Freshman Non Physics Major are chosen:
Prob (Jr No-Ph Mjr) = Jr No-Ph Mjr / Total
= 18 / 82 = 0.2195
Prob (Fr No-Ph Mjr) = Fr No-Ph Mjr / Total (remaining)
= 9 / 81 = 0.1111
Prob [ Jr No-Ph Mjr & Fr No-Ph Mjr ] = 0.2195 x 0.1111 = 0.02439
≈ 0.0244
Please help me with this
Answer:
7 Newton's East
Explanation:
when the force is going in the same direction in this case east, you add the forces.
Why does the brightness of a bulb not change noticeably when you use longer copper wires to connect it to the battery?
a. All the current is used up in the bulb, so the connecting wires don't matter.
b. Very little energy is dissipated in the thick connecting wires.
c. Electric field in the connecting wires is zero, so emf = E_bulb * L_bulb.
d. Current in the connecting wires is smaller than current in the bulb.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
Answer:
Options B & E are correct
Explanation:
Looking at all the options, B & E are the correct ones.
Option B is correct because the thicker the wire per unit length, the lesser resistance it will posses and the lesser the energy that will be dissipated by the wire and in return more energy will be dissipated by the bulb.
Option E is also correct because the resistance of the copper wires is low enough to ensure that there's not much drop in voltage across the copper wires. Thus, there will not be any noticeable differences in the voltage across the bulb.
Option A is not correct because the current is not used up and thus the charge is conserved, and it will circulate just through the circuit.
Option C is not correct because although the Electric field along the wire is not zero, it is very small.
Option D is not correct because the wires and the light bulb are connected in series and as such, the current in both the wires and the light bulb will be identical.
The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is :
b. Very little energy is dissipated in the thick connecting wires.
e. The electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
"Energy"The brightness of a bulb that not change noticeably when you use longer copper wires to connect it to the battery is very little energy is dissipated in the thick connecting wires and the electric field in connecting wires is very small, so emf almost = E_ bulb * L_bulb.
The thicker the wire per unit length, the lesser resistance it'll posses and the lesser the vitality that will be scattered by the wire and in return more vitality will be disseminated by the bulb.
The resistance of the copper wires is low sufficient to guarantee that there's not much drop in voltage over the copper wires. Hence, there will not be any noticeable contrasts within the voltage over the bulb.
Thus, the correct answer is B and E.
Learn more about "Circuit":
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What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equation is [tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
The value of c is [tex]c = 2.998 *10^{8} \ m/s [/tex]
Explanation:
From the question we are told that
Generally the equation that relates the speed of light to other fundamental electromagnetic constants is
[tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
Here c is the speed of light
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
and [tex]\epsilon_o[/tex] is the permittivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ C/V \cdot m[/tex]
So
[tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7} * 8.85*10^{-12}} }[/tex]
=> [tex]c = 2.998 *10^{8} \ m/s [/tex]
A soccer player kicking a ball; the ball soaring through the air and landing on the ground
If you are driving to see your cousins at a speed of 84.6 km/h and it took you 6.5 h to get there, how far did you travel?
Answer: 549.9 km
Explanation: 84.6km every hour so 84.6*6.5= 549.9
7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
how can philosophy help you become a productive citizen
Answer:
Philosophy is a study that involves the nature of knowledge and truth. It serves as a guide that helps an individual seek which things are valuable and essential in life. ... It gives you a sense of direction, knowing the weight of things, therefore making you more productive.
A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.
Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
You are driving your car at 45 m/s, when a raccoon runs into the street in front of you. You slam on the brakes and come to a stop in 5 seconds. What is the acceleration of your car?
Answer:
-9m/s²
Explanation:
Given parameters:
Initial velocity = 45m/s
Final velocity = 0
duration = 5s
Unknown:
acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time;
Acceleration = [tex]\frac{v- u}{t}[/tex]
v is the final velocity
u is the initial velocity
t is the time taken
Input the parameters and solve;
Acceleration = [tex]\frac{0 - 45}{5}[/tex] = -9m/s²
The car accelerates at a rate of -9m/s² which is a deceleration
A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?
Answer:
578.66 N
Explanation:
The first step is to calculate the mass
mgh= 3200J
3200/9.8×5.53
3200/54.194
m = 59.047 kg
Therefore the weight can be calculated as follows
Weight = m × g
= 59.047 × 9.8
= 578.66 N