Answer:
a) 2.8 m/s2
b) 2.2 m/s2
Explanation:
The lecture on the bathroom scale is equal to the normal force that the scale exerts on you, and is directed upward.There is another force acting on you, which is gravity.This force (which we call weight) is the product of the mass times the acceleration due to gravity, g, and is directed downward, opposite to the normal force.When the elevator is at rest, no acceleration takes place, so according Newton's 2nd Law, no net force must be exerted on you.In this condition, the normal force Fn₀ must be equal to the weight:[tex]F_{no} = m * g = 177 lb (1)[/tex]Since the normal force takes any value needed to satisfy Newton's 2nd law, the two extreme lectures can be expressed as follows, in terms of the two forces acting on you while the elevator is moving upward:[tex]F_{n1} = m * (g +a_{1}) = 227 lb (2)[/tex]
[tex]F_{n2} = m * (g +a_{2}) = 138 lb (3)[/tex]
Replacing m by 177lb/g (a given), and rearranging, we can solve (1) for a₁, as follows:[tex]a_{1} = (\frac{227lb}{177lb} *g) - g = 0.28*g = 0.28*9.8m/s2 = 2.8 m/s2 (4)[/tex]
As it can be seen, the normal force takes a larger value in order to be compliant with the upward acceleration that opposes to gravity.In the same way, we can find the magnitude of the downward acceleration when the elevator brakes to a stop, from (3):[tex]a_{2} = (\frac{138lb}{177lb} *g) - g = -0.22*g =- 0.22*9.8m/s2 = -2.2 m/s2 (5)[/tex]
In this case, the normal force takes a lower value than at rest, due to the acceleration has the same direction as gravity.How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then
pushing it back to its original position? How much power is used if this work is done in 20 sec?
Answer:
35/2 J/s
Explanation:
Just use the 2 formulas
Work done = Force * distance moved
Power = Work done/time
WD = 7 * 50 = 350
Power = 350 / 20
= 35/2 J/s
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
Fill in the graph for 50 points
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Explanation:
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Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
At the end of the previous experiment, aclumsy scientist drops the coil, while still in the magnetic field, and still oriented with its plane perpendicular to the magnetic field, denting it and changing its shape to a semi-circle. The new shape has the same perimeter, but a different area, and it takes 0.036s to deform. What isthe average induced EMF during this mishap
Answer:
hello your question has some missing parts below are the missing parts
A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.
answer : 1 volt
Explanation:
Determine the Average induced EMF during this mishap
A' = A/2 ( for a semi circle )
where A = [tex]\frac{\pi r^2}{2}[/tex]
To determine the Average induced EMF apply the relation below
| E | = η * [tex]\frac{\beta A}{T}[/tex] ----- ( 1 )
Replace A in equation 1 with A = [tex]\frac{\pi r^2}{2}[/tex]
hence equation becomes : | E | = η * βπr^2 / 2T'
where : T' = 0.0365 , β = 0.2 , η = 10 , r = 0.107
∴| E | = 0.999 ≈ 1volts
2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:
a.On Earth, where the acceleration due to gravity is 9.8m/s?
b. On Mars, where the acceleration due to gravity is 3.8m/s?
Answer:
a=24.5 b=9.5
Explanation:
30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Answer:
[tex]18.18\ \text{m/s}[/tex]
[tex]6.82\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of large object = 8 kg
[tex]m_2[/tex] = Mass of smaller object = 3 kg
When large mass is moving
[tex]u_1[/tex] = 25 m/s
[tex]u_2[/tex] = 0
For completely inelastic collision we have the relation
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}[/tex]
Speed of the combined mass when the larger object is moving is [tex]18.18\ \text{m/s}[/tex]
When smaller mass is moving
[tex]u_1[/tex] = 0
[tex]u_2[/tex] = 25 m/s
[tex]v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}[/tex]
Speed of the combined mass when the smaller object is moving is [tex]6.82\ \text{m/s}[/tex]
Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)
Answer:
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]
Explanation:
This is an interesting problem, no data is given, so the result is a general expression.
Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂
we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved
initial instant. Just before the crash
L₀ = L₁ + L₂
with
L₁ = I₁ w₁
the moment of inertia of a disc with an axis passing through its center is
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂
final instant. Right after the crash
L_f = I w
in angular momentum it is a scalar quantity, so it is additive
I = I₁ + I₂
angular momentum is conserved
L₀ = L_f
I₁ w₁ + I₂ w₂ = I w
w = [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex] (1)
We already have the angular velocities of the system, let's find the kinetic energy of it
initial
K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²
final
K_f = K = ½ I w²
the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses
ΔK = K_f -K₀
ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²) (2)
In this chaos we know all the values for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated
Now if all this variation of energy turns into heat
Q = ΔK
m_{total} c_e ΔT = ΔK
where the specific heat of the bear discs must be known, suppose they are of the same material
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex] (3)
to make a special case, we suppose some data
the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC
we look for the angular velocity
I₁ = I₂ = I₀
I = 2 I₀
we substitute in 1
w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io
w = w₁ /2
we look for the variation of the kinetic energy with 2
ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)
ΔK = ¼ I₀ w₁² -½ I₀ w₁²
ΔK = - ¼ I₀ w₁²
the negative sign indicates that the kinetic energy decreases
We look for the change in Temperature with the expression 3
ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce
ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]
ΔT = [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]
ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]
in this expression all the terms are contained
The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
What is internal energy?The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.
The given data in the problem is;
[tex]\rm \omega_1[/tex] is the angular velocity of disk 1
[tex]\rm \omega_2[/tex] is the angular velocity of disk 2
r₁ is the radius of disk 1
r₂ is the radius of disk 2
m₁ is the mass of disk 1
m₂ is the mass of disk 2
Momentum before the collision;
[tex]\rm L_1 = I_1 \omega_1[/tex]
The moment of inertia of disc 1
[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]
The momentum gets conserved;
[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]
The change in the kinetic energy is;
[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]
The change in the energy gets converted into heat;
[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]
The change in the temperature is
[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]
The internal energy change is found by;
[tex]\rm \triangle E = mc_v dt[/tex]
[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]
Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
To learn more about the internal energy refer to the link;
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Tom has a mass of 50,000 g and runs up a flight of stairs 4 m high in 12.5 seconds.
Calculate Tom’s power. (g = 10 m/s2)
Answer:
160 watts.
Explanation:
Remark
Power = Work / Time
Work = F * d
Note: Since he is running up stairs he is doing work against gravity.
Givens
m = 50000 g kg / 1000 grmsm = 50000 / 1000 = 50 kgh = 4 mtime = 12.5 secondsg = 10 m/s^2Formula
P = W * d/tW = m*g *d / tSolution
P = 50kg * 10 m/s^2 * 4 m / 12.5 P = 160 watts.
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
BRAINLEST FOR CORRECT ANSWER PLEASE
Which has more momentum: a 3 kg sledgehammer swung at 1.5 m/s OR a 4 kg sledgehammer swung at 0.9 m/s? SHOW YOUR WORK
Answer:
Sledgehammer A has more momentum
Explanation:
Given:
Mass of Sledgehammer A = 3 Kg
Swing speed = 1.5 m/s
Mass of Sledgehammer B = 4 Kg
Swing speed = 0.9 m/s
Find:
More momentum
Computation:
Momentum = mv
Momentum sledgehammer A = 3 x 1.5
Momentum sledgehammer A = 4.5 kg⋅m/s
Momentum sledgehammer B = 4 x 0.9
Momentum sledgehammer B = 3.6 kg⋅m/s
Sledgehammer A has more momentum
Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle.
Answer:
The amount of torque is 0.03 N.m.
Explanation:
To find the amount of torque we need to use the following equation:
[tex] \tau = \vec {r} \times \vec{F} = rFsin(\theta) [/tex] (1)
Where:
r: is the radius = 1x10⁻² m
F: is the force = 6 N
θ: is the angle = 30°
By entering the above values into equation (1) we have:
[tex]\tau = 1 \cdot 10^{-2} m*6 N*sin(30) = 0.03 N.m[/tex]
Therefore, the amount of torque is 0.03 N.m.
I hope it helps you!
Match the descriptions with the graphs !
Answer:
Graph 1 matches with B, 2 with A, and 3 with C.
Explanation:
Graph 2 shows a car whose distance part of the graph is not going up or down, while the time going up. That means that the car is stopped. Graph 1 shows a straight line, meaning that the car is traveling at a constant speed. Graph 3 is a curved line, meaning the speed of the car is changing somehow, and since the line is becoming more horizontal, the car is getting slower.
Which of the following would produce the most power?
b
ОООО
A mass of 10 kilograms lifted 10 meters in 10 seconds
A mass of 5 kilograms lifted 10 meters in 5 seconds
A mass of 10 kilograms lifted 10 meters in 5 seconds
A mass of 5 kilograms lifted 5 meters in 10 seconds
d
Answer:
A mass of 10 kilograms lifted 10 meters in 5 seconds.
Explanation:
Power can be defined as the energy required to do work per unit time.
Mathematically, it is given by the formula;
[tex] Power = \frac {Energy}{time} [/tex]
But Energy = mgh
Substituting into the equation, we have
[tex] Power = \frac {mgh}{time} [/tex]
Given the following data;
Mass = 10kg
Height = 10m
Time = 5 seconds
We know that acceleration due to gravity is equal to 9.8 m/s²
[tex] Power = \frac {10*9.8*10}{5} = 490 Watts [/tex]
Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.
ocean currents are always cold true or false
a forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite direction determine the magnitude and direction of the resultant force acting on the crate
Answer:
130n on the 2nd horizontal
Explanation:
A 744 N force is applied to an object to reach an acceleration of 24 m/s2. What is the objects mass?
31kg
Explanation:
F = ma
m = F/a
m = 744N/24m/s^2
m = 31kg
(*Newton's Second Law*)
If you wrap 150 coils of heavy wire around a big iron nail and attach the ends of the wire to a 6.0v battery, you have a A) radio B) electromagnet C) galvanometer D) ammeter
Answer:
B
Explanation:
Because of the voltage attached to the iron nail
A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebble of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare
Hello,
QUESTION)✔ We have: KE = PE (potential energy)
PE = m x g x h
The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2
PE1 = PE2 ⇔ PE1/PE2 = 1
[tex]\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4[/tex]
The mass m1 is therefore 4 times greater than that of the stone of mass m2.
True or false? A system must contain more than one object.
Answer:
true
Explanation:
normally -No system has ever performed well with one object.
A system must contain more than one object is a true statement.
What is system?A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.
Normally, no system has ever performed well with one object.
To learn more about System here
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which two options describes behaviors of particles that are related to the chemical properties of the materials
a- forming hydrogen bonds between them
b- reacting quickly with water
c- having a high mass
d- forming bonds with other atoms
Answer:
The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.
Explanation:
I took the quiz on a.pex and these were correct.