Answer:
A.Radio waves
Explanation:
ocean currents are always cold true or false
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.
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]
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:
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
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
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
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.
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].
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)]
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.
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
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
https://brainly.com/question/24893867
#SPJ2
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.
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.
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
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:
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!
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*)
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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.