Answer:
The initial velocity of all rocket is zero.Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R = I/V D.) R = IV
Answer:
[tex]V = IR[/tex]
Explanation:
Required
Which equation represents ohm's law?
Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).
Mathematically, this can be represented as:
[tex]I\ \alpha\ \frac{V}{R}[/tex]
Convert the expression to an equation
[tex]I\ =\ \frac{V}{R}[/tex]
Multiply both sides by R to make V the subject
[tex]I\ * R\ =\ \frac{V}{R} * R[/tex]
[tex]I\ * R\ =V[/tex]
Reorder
[tex]V = I\ * R[/tex]
[tex]V = IR[/tex]
Option (a) is correct; Others are not
Answer:
V=ir
Explanation:
Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR. The correct answers from the choices are:
v = ir
Two ice-skaters are skating in circles on a frozen pond. Maria is making large circles with a radius of 12 m and skating 4.5 m/s. Her friend, Samantha, is making smaller circles with a radius of 6 m but is not skating as quickly, going only 3.8 m/s. ii. How could each skater increase her centripetal acceleration without changing the size of her path? Explain your reasoning.
Answer:
v_maria > 4.5 m/s
v_samantha > 3.8 m/s
Explanation:
Formula for centripetal acceleration is;
a_c = v²/r
Where;
v is speed
r is radius
For Maria;
r = 12 m
v = 4.5 m/s
Thus;
a_c = 4.5²/12
a_c = 1.6875 m/s²
For Samantha;
r = 6
v = 3.8 m/s
Thus;
a_c = 3.8²/6
a_c = 2.41 m/s²
We want to find how could each skater increase her centripetal acceleration without changing the size of her path.
From the centripetal acceleration formula, since the size of path can't be changed it means the radius can't be changed and so the only thing that can now increase the centripetal acceleration is when the speed increases.
Thus;
For Maria, she has to move with a faster speed. Thus: v_maria > 4.5 m/s
For Samantha, she also has to move with a faster speed. Thus; v_samantha > 3.8 m/s
A scientist who wants to study the affects of fertilizer on plants sets up an experiment. Plant A gets no fertilizer, Plant B gets 5 milligrams of fertilizer, and Plant C gets 10 milligrams of fertilizer. Which plant is the control group in this experiment?
Answer: Plant A
Explanation: The control group is participants who do not receive the experimental treatment.
The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How
high will the water spray into the air?
Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Paco pulls a 67 kg crate with 738 N and of force across a frictionless floor 9.0 M how much work does he do in moving the crate
Answer:
W = 6642 J
Explanation:
Given that,
Mass of a crate, m = 67 kg
Force with which the crate is pulled, F = 738 N
It is moved 9 m across a frictionless floor
We need to find the work done in moving the crate. Let the work done is W. It is given by :
W = F d
W = 738 N × 9 m
= 6642 J
So, the work done is 6642 J.
on the moon which object would fall with the same acceleration?
Answer:
Since there is no air resistance on the Moon, all objects would be in free fall at 1.6 m/s2. This means that they would all hit the ground at the same time if released simultaneously from the same height, but at a slower speed compared to objects free falling in a vacuum on Earth.
Explanation:
Item 2
An object is transported to three different planets in the solar system.
Which statement is true about that object?
The object’s mass and weight stay the same.
The object’s mass changes, but its weight stays the same.
The object’s mass and weight both change.
The object’s weight changes, but its mass stays the same.
Answer:
the object's weight changes but its mass stays the same
Answer:
the object's weight changes but its mass stays the same
Explanation:
100% correct
Which change in an object would increase the force needed to move the object?
А.
decreasing the velocity of an object
B
increasing the volume of an object
с
decreasing the mass of an object
D
increasing the mass of an object
Answer:
D i think
Explanation:
Heavier objects (objects with more mass) are more difficult to move and stop.
Answer:
Increasing the mass of the object (option D in the list of answers)
Explanation:
Recall that F = m x a
therefore, if the mass increases, the force increases
Object A has a mass of 2 kg; object B’s mass is 10 kg. If object B is at rest and object A runs into it at low speed, object B:
A. Will move off at high speed.
B. Cannot move off with high speed.
C. Will loose speed in the collision.
D. Will remain at rest.
Answer:
Cannot move off with high speed
Explanation:
Just did it
Changing which factor would NOT have an influence on the kinetic energy of a moving van loaded with 100 kg bags mulch of with a total mass of 1,500 kg. The vehicle is traveling across an open parking lot at a speed of 5 m/s.
Question 8 options:
A: direction the moving van is going across the parking lot.
B: increasing the rate of speed without altering the mass of the vehicle or its contents.
C: emptying the moving van one bag at a time at a constant rate.
D: adding more contents (increasing the overall mass) to the van while the van is in motion traveling 5 m
Answer:
A. The direction
Explanation:
I did the test lol
Answer:
A
Explanation:
i took the test
A force Ě = F, î + Fy h acts on a particle
that undergoes a displacement of g = s, it
Sy Ĵ where F, = 10 N, F, = -1 N, 8z = 4 m,
and sy=1 m.
Find the work done by the force on the
particle.
Answer in units of J.
Answer:
Pt 1: [tex]W=39J[/tex]
Pt 2: θ = [tex]19.7[/tex]
Explanation:
Part 1:
[tex]W=FS[/tex]
[tex]W=F_{_xs_x}+F{_ys_y}[/tex]
imagine the "i"and the "j" with the hat
[tex]W=(10i-1j)(4i+3j)[/tex]
[tex]W=(10*4)-(1*1)[/tex]
[tex]W=40-1[/tex]
[tex]W=39J[/tex]
Part 2:
[tex]|F|=\sqrt{10^2+(-1)^2}[/tex]
[tex]|F|=\sqrt{101}[/tex]
[tex]|S|=\sqrt{4^2+1^2}[/tex]
[tex]|S|=\sqrt{17}[/tex]
[tex]W=|F||S|cos[/tex]θ
[tex]39=(\sqrt{101})(\sqrt{17} )cos[/tex]θ
θ = [tex]19.7[/tex]
Hope it helps
How often does the World Cup happen?
Your answer
Answer:
every 4 years
Explanation:
Answer:
Every four years
Explanation:
A 5.0 kg block is pushed 2.0 m at a con-
stant velocity up a vertical wall by a constant
force applied at an angle of 30.0° with the
horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s2.
F
30°
2 m
5 kg
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.
Answer:
[tex]W_F=127.64283 J[/tex]
Explanation:
Information Given:
[tex]m = 5kg[/tex] [tex]v=constant[/tex]
Key: μ = Kinetic Friction (Kf) θ = Theta α = 180° N = Normal Force
[tex]W_F=F_ydcos[/tex]θ
[tex]W_F=Fdsin[/tex]θ
[tex]_{net}F_y = sin[/tex]θ-μ[tex]N-mg=0[/tex]
[tex]_{net}F_x = 0[/tex]
[tex]N=Fcos[/tex]θ
[tex]Fsin[/tex]θ-μ[tex]N=mg[/tex]
[tex]Fsin[/tex]θ-μ[tex]Fcos[/tex]θ[tex]=mg[/tex]
[tex]F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex] →[tex]W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex]
[tex]W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}[/tex]
[tex]W_F=127.64283 J[/tex]
At the instant its angular displacement is 0.32 rad, the angular acceleration of a physical pendulum is -630 rad/s2. What is its angular frequency of oscillation?
6.6 rad/s
14 rad/s
20 rad/s
44 rad/s
200 rad/s
Answer:
200rad/s
Explanation:
The angular frequency of oscillation of the pendulum is 44.3 rad/s.
What is meant by angular frequency ?Angular frequency is defined as the rate of change of angular displacement in a simple harmonic motion.
Here,
Angular displacement of the pendulum, θ = 0.32 rad
Angular acceleration, α = -630 rad/s²
We know that the equation for angular acceleration is given by,
α = -ω²θ
where ω is the angular frequency
ω² = -α/θ
ω² = 630/0.32
ω² = 1968.8
Therefore,
Angular frequency, ω = √1968.8
ω = 44.3 rad/s
Hence,
The angular frequency of oscillation of the pendulum is 44.3 rad/s.
To learn more about angular frequency, click:
https://brainly.com/question/14103673
#SPJ3
what is the need of force in our life
Answer:
A force is a push or a pull and it affects our daily lives because without force,people would not be able to open and close stuff or lift up our arms or legs .....or anything, for that matter.
Explanation:
brainly
A 100 kg man stands still. Gravity pushes on him with an acceleration of 9.8 m/s^2. What it the force the man feels from gravity? *
0 N
98 N
90 N
980 N
Answer:
980 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 100 × 9.8
We have the final answer as
980 NHope this helps you
What is the minimum work done by the heart to pump 130 g of blood from a person's foot to his heart 1.5 m away?
A. 0J.B. 4.7J.C. .47J.D. 4700J.
Answer:
The minimum work done by the heart is 1.911 J
Explanation:
Given;
mass of the blood, m = 130 g = 0.13 kg
height the blood is pumped, h = 1.5 m
Apply work energy theorem, the gravitational potential energy of the height in which the blood is pumped to, is equal to the work done in pumping the blood form the foot to the heart.
E = mgh = W
where;
g is acceleration due to gravity = 9.8 m/s²
W = (0.13)(9.8)(1.5)
W = 1.911 J
Thus, the minimum work done by the heart is 1.911 J
Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the circuit’s current in Amps according to the following table.
Trial Number
Voltage (V)
Resistance (Ω)
Current (A)
1
1.5
200
2
1.5
100
3
3.0
200
4
3.0
100
For which trial would the students measure the smallest current in the circuit? (AKS 10a)
A.
Trial 1
B.
Trial 2
C.
Trial 3
D.
Trial 4
Answer:
bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon
Explanation:
bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo
what is the portion of an electric circuit that is being powered
Answer:
Component
Explanation:
All circuits have some basic parts, called components. One component is the power source, also called a voltage source. The power source is what pushes the electricity through the circuit.
How much heat is absorbed by 3 kg honey baked ham as energy from the oven causes its temperature to change from 10°C to 60°C? (Specific heat capacity of ham 3287 cal/kgºC)?
Answer:
The answer is 493.05kJ
Explanation:
given
mass m= 3kg
t1= 10°C
t2= 60°C
Specific heat capacity of ham =3287 cal/kgºC
Required,
The quantity of heat, Q
Step two:
the formula is
Q= mcΔT
substitute
Q= 3*3287(60-10)
Q=9861*50
Q=493050 J
Q= 493.05kJ
Under state law, what is the blood-alcohol limit for legally operating a motor vehicle?
Answer:
HOPE IT HELPS.....
Explanation:
The legal standard for drunkenness across the United States ranges from 0.10 to 0.08. If a person's BAC measures 0.08, it means that there are 0.08 grams (i.e., 80 mg) of alcohol per 100 ml of blood. The American Medical Association says that a person can become impaired when the blood alcohol level hits 0.05.
Answer: 0.08
Driving While Intoxicated or Impaired is Illegal
Under Michigan law, it is illegal to drive: While intoxicated, or impaired, by alcohol, controlled substance, or other intoxicating substance. With a bodily alcohol content of 0.08 or more. (This crime is one of the driving while intoxicated offenses.)
Hope this helps... Stay safe and have a Merry Christmas!!!!!!!! :D
_____ are group of tissue working together to perform a certain job.
Answer:
organ
Explanation:
Answer:
an organ
Explanation:
cell -> tissue -> organ -> organ system -> organism
A 60-kg jogger runs up a long flight of
stairs in 4.0 s. The vertical height of the
stairs is 4.5 m.
a. Estimate the jogger's power
output in watts and horsepower.
b. How much energy did this
require?
Explanation:
Given parameters:
Mass of Jogger = 60kg
Time = 4s
Vertical height = 4.5m
Unknown:
Jogger's power output = ?
Energy required = ?
Solution:
The power output of the jogger is defined as the rate at which work is done.
Power = [tex]\frac{force x distance}{time}[/tex]
Now insert the parameters and solve;
Work done = Force x distance = mgh
m is the mass
g is the acceleration due to gravity = 9.8m/s²
h is the height
Work done = 60 x 9.8 x 4.5 = 2646J
Power = [tex]\frac{2646}{4}[/tex] = 661.5W
Energy required;
The work done here is also the energy required;
Energy required = 2646J
The energy required is 2646 J and the jogger's power output 661.5 W.
Given here,
Mass of Jogger = 60 kg
Time = 4s
Vertical height = 4.5 m
The power output of the jogger is defined as the rate of energy tranfer work is done.
Power = W/t
Where,
W- work = mgh = 60 kg x 9.8 m/s² x 4.5 m = 2646 J
t - time - 4 s
Put the values, we get
Power = 661.5W
To know more about power,
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Introduction to Simple Machines
This activity will help you meet this educational goal:
You will compare and contrast information from a video with information from a text.
Directions
Read the instructions for this self-checked activity. Type in your response to each question, and check your answers. At the end of the activity, write a brief evaluation of your work.
Activity
Watch this video and then answer the following questions based on what you learned.
Part A
How does a bicycle make work easier?
Part B
Which two examples of levers are mentioned in the video?
The picture shows a bicycle’s pedals. Look at the shaft that the pedals are attached to. Do you think the shaft is a lever? Why or why not?
Answer:
word for word answers!
Explanation:
1) Part A: By pedaling a bicycle lightly, the rider can go a long way
2) Part B: The two examples mentioned in the video are the handlebars and the brakes
3) Yes, it’s a type of lever because the two pedals rotate around a fixed point
Which kind of lens can correct a myopic (nearsighted) eye?
Answer:
diverging lens
Explanation:
Since the nature of the problem of nearsightedness is that the light is focused in front of the retina, a diverging lens will serve to diverge light before it reaches the eye. This light will then be converged by the cornea and lens to produce an image on the retina.
Cells work ALONE and do not depend on each other to function. True or False? hELPpPP plSSSS
Answer:
False
Explanation:
Answer:
False because cells depend on the others and absolutely positively CANNOT work alone.
Explanation:
When liquid water gets into cracks of rock and freezes, it __ and ___.
If the moon crossed the meridian at midnight what phase must the moon be in?
Answer:
Waxing Gibbous
Explanation:
New Moon Rises at sunrise, transits meridian at noon, sets at sunset
Explanation:
Gregor Mendel observed that pea plant traits did not blend in their offspring?
A string with mass per unit length of 0.003 kg/m is plucked with an amplitude of 2 cm. The string is 30 cm long, and under a tension of 30 newtons. What are the frequencies of the first 5 harmonics?
Answer:
First harmonics = 333.33N
Second harmonics = 500.01N
Third harmonics = 666.68N
Fourth harmonics = 833.35N
Fifth harmonics = 1000.02N
Explanation:
The formula for calculating the fundamental frequency in string is expressed as;
[tex]F_0 = \frac{1}{2L}\sqrt{\frac{T}{m} }[/tex] where;
L is the length of the string = 30cm = 0.3m
T is the tension in the string = 30N
m is the mass per unit length of the string = 0.003kg/m
Get the fundamental frequency first by substituting the given values into the formula;
[tex]F_0 = \frac{1}{2(0.3)}\sqrt{\frac{30}{0.003} }\\F_0 = \frac{1}{0.6}\sqrt{10,000}}\\F_0 = \frac{1}{0.6} * 100\\[/tex]
F0 = 166.67N
Harmonics are the integral multiples of the fundamental frequency.
First harmonics F1 = 2F0 = 2(166.67) = 333.33N
Second harmonics F2 = 3F0 = 3(166.67) = 500.01N
Third harmonics F3 = 4F0 = 4(166.67) = 666.68N
Fourth harmonics F4 = 5f0 = 5(166.67) = 833.35N
Fifth harmonics F5 = 6f0 = 6(166.67) = 1000.02N