Answers
Answer:
Your ans is B)
Explanation:
Hope its right
if wrong firgive me
Related Questions
PLEASE HELP ME WITH ONE QUESTION
If 1000 J of heat energy is lost from a piece of hot metal which is placed into a cup of cool water, how much energy is gained by the water?
Answers
Answer:
1000J
Explanation:
According to the law of calorimetry;
Heat lost by cold substance = Heat gained by hot substance
If 1000 J of heat energy is lost from a piece of hot metal,
heat lost by hot substance = 1000J
Since we are to determine how much energy is gained by the water, using the law
Heat lost by hot metal = heat gained by the water
Hence heat gained by the water = 1000J
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you can plz
Answers
Answer:
P = 11666.6 W
Explanation:
Given that,
Work done by the motor, W = 3500 kJ
Time, t = 5 min = 300 s
We need to find the power developed by the motor. Power developed is given by :
[tex]P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W[/tex]
So, the required power is 11666.6 W.
Direction of Wave Travel
С
Which letter
correctly
identifies the
wavelength of
this wave?
А
B
A. A
B. B
C. C
Answers
Answer:
I THINK it's A but I'm not completely sure
A projectile is thrown with a velocity of 40 m/s, 45 degrees above the +x-axis. Determine its initial x-velocity and its initial y-velocity.
Answers
x-component is V·cos(angle)
y-component is V·sin(angle)
When the angle is 45°, its sin and cos are both 1/2·√2 .
Vx = 20·√2 = 28.28 m/s
Vy = 20·√2 = 28.28 m/s
A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
Answers
Answer:
The correct answer is "12 rad/s"
Explanation:
The given values are,
Mass of rod,
M = 140 g
i.e.,
= 0.14 kg
Length,
L = 57 cm
i.e.,
= 0.57 m
Mass of beads,
M = 30 g
i.e.,
= 0.03 kg
Angular speed,
r = 11 cm
i.e.,
= 0.11 m
Now,
The inertia of rods will be:
= [tex]\frac{1}{12}ML ^2[/tex]
On substituting the values, we get
= [tex]\frac{1}{12}\times 0.14\times (0.57)^2[/tex]
= [tex]0.0037905 \ kg-m^2[/tex]
The inertia of beads will be:
= [tex]mr^2[/tex]
On substituting the values, we get
= [tex]0.03\times (0.11)^2[/tex]
= [tex]0.000726 \ kg-m^2[/tex]
The total inertia will be:
= [tex]Inertia \ of \ rods+Inertia \ of \ beads[/tex]
= [tex]0.0037905 + 0.000726[/tex]
= [tex]0.0045165 \ kg-m^2[/tex]
now,
The angular speed of the system will be:
⇒ [tex]L_1w_1=L_2w_2[/tex]
On substituting the values in the above equation, we get
⇒ [tex]0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2[/tex]
⇒ [tex]0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2[/tex]
⇒ [tex]w_2 = 12 \ rad/s[/tex]
A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a hill of unknown height. It then goes down that first hill and up another circularly shaped hill of height 10 meters and radius 15 meters. The ride is designed so that the cart just barely loses contact with the track at the top of the second hill (10 meters above ground level). What is the height of the starting hill relative to ground
Answers
Answer:
hinit = 17.5 m
Explanation:
Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:[tex]E_{o} = U_{o} = m*g*h_{init} (1)[/tex]
When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.We can express this final energy as follows:[tex]E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)[/tex]
In order to find hinit, we need to make (1) equal to (2), and solve for it.In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).We know also that there is a force that keeps the car along the circular track, which is the centripetal force.This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.So, the centripetal force must be equal to the weight.The centripetal force can be expressed as follows:[tex]F_{c} = m*\frac{v_{f} ^{2}}{R} (3)[/tex]
We have just said that (3) must be equal to the weight:[tex]F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)[/tex]
Simplifying, and rearranging, we can solve for vf², as follows:[tex]v_{f}^{2} = R*g (5)[/tex]
Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:[tex]h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)[/tex]