Answer
Sun and Moon's gravity
Question 3
Which of the following is an example of positive acceleration?
A car increases its speed from 5 m/s to 10 m/s.
A car is traveling at 25 m/s for 5 minutes.
A car is traveling at 25 m/s and begins to slow down as it approaches a stop sign.
A car is in park.
Answer: 1ST OPTION A car increases its speed from 5 m/s to 10 m/s
Explanation:IN 1ST OPTION : because the vel and force are in the same direction therefore acceleration is positive
IN 2ND & 3RD OPTION : the force and velocity in opposite direction so its velocity decreases therefore acceleration is negitive
What does it mean to discover
the truth in the context of
detective work? What kinds of
truth can be discovered?
Answer:
When the machine indicates a lie, what is the probability that the suspect is really lying? If the machine does not indicate a lie, what is the probability that the suspect is really telling the truth?
My solution if the probability that a test subject is chosen at random from the 200 in fact lie = 80/(80+5)=0.94
Then when 15% of people arrested lie, the probability that the suspect is really lying is 0.15*0.94=0.141
What is the magnitude of the electric field strength at a position that is 1.2 m from a point charge of 4.2 \times 10^{-6} \mathrm{C}4.2×10
−6C?
Answer:
lil durk a goat
Explanation:
What is the reading of the Voltmeter shown in the figure?
a. 8V
b. 10V
c. 16V
d. 20V
Answer:
Correct option is A)
Let us assume voltmeter is ideal:-It means voltmeter has infinite resistance and current will not flow in voltmeter.
R
eq
in upper branch = R
eq
in lower branch. So current will get equally divide into both branches.
V
A
−V
B
=4×1(1)
V
A
−V
D
=16×1(2)
Subtract 2 and 1
V
B
−V
D
=12V
A voltmeter will read 12 V
A wheel rolls 5 revolutions on a horizontal surface without slipping. If the center of the wheel moves 3.2 m, what is the radius of the wheel
The radius of the wheel is 0.102 m.
What is radius of a circle?The radius of a circle is a straight line that comes from the center of a circle to any point of the circumference of the circle.
To calculate the radius of 5 revolutions of the wheel, we use the formula below.
Formula:
d = 10πr.............. Equation 1Where:
d = Total distance moved by the wheelr = radius of the wheel.π = pieMake r the subject of the equation
r = d/(10π)........... Equation 2From the question,
Given:
d = 3.2 mπ = 3.14Substitute these values into equation 2
r = 3.2/(10×3.14)r = 3.2/31.4r = 0.102 m.Hence, the radius of the wheel is 0.102 m.
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I need help algebra in physics
Answer:
A: 0.26
Explanation:
using my friend account again
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Which example is a simple machine?
1. ramp
2. bicycle
3. car
4. steering wheel system of a boat
Answer:
Ramp
Explanation:
All of the other things are fairly complex, bikes have to be made with multiple parts so do cars and steering wheels are another part of a boat, a ramp usually consists of two or fewer parts making it the simplest.
Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain how the following will be different for car 2: a) braking distance, b) thinking distance and c) stopping distance.
Answer: car 1 is going how fast
Explanation: no need to answer without speed I won't know distance.
Two spheres carry a charge of 1x10^-7 C. Their centers are separated by 0.7m. Calculate the magnitude of the force between the two charged objects.
Write your answer in scientific notation using one significant figure
PLEASE HELP ME
We have that the force between the two charged objects is mathematically given as
F=2.349N
Magnitude of the force between two charged objects.Question Parameters:
Two spheres carry a charge of 1x10^-7 C.
Their centers are separated by 0.7m.
Generally the equation for the Force is mathematically given as
[tex]F=\frac{q1q2}{4*\pi e*r}[/tex]
Therefore
[tex]F=\frac{(1e-7)^2}{9e9*0.7}[/tex]
F=2.349N
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A ball with a weight of 15 N is thrown striaght up into the air with an initial speed of 14 m/s. What is the maximum height it will achieve above its starting position
Answer:
10m above initial height
Explanation:
let acceleration due to gravity = 9.8 m/s/s
v final = v initial - time • gravity
0 = 14m/s - t • 9.8m/s/s
t = 1.43 seconds to reach max height
distance = 1/2 • acceleration • time^2
distance = 1/2 • 9.8 • 1.43^2
distance = 10 meters above initial height
A 10 kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 5 N opposing the motion.
Calculate the acceleration of the object.
Newton second's law
∑F = m.a
40 - 5 = 10.a
35 = 10.a
a = 3.5 m/s²
Answer:
The acceleration of the object was 3.5 m/s².
Explanation:
Let the direction of the horizontal force be positive.
By Newton's second law, the sum of the horizontal forces is as follows:
[tex]\displaystyle \sum F_x = F_A - F_k = ma[/tex]
Where Fk is the frictional force and FA is the horizontal force.
Substitute in known values and solve for acceleration a:
[tex]\displaystyle \begin{aligned} (40\text{ N}) - (5\text{ N}) & = (10\text{ kg})a \\ \\ a &= 3.5\text{ m/s$^2$}\end{aligned}[/tex]
Hence, the acceleration of the object was 3.5 m/s².
A new company is making security cameras. These security cameras will be different than other security cameras because instead of detecting motion, they will detect body heat.
What type of electromagnetic wave will the camera detect?
A
gamma ray
B
infrared
C
microwave
D
radio wave
Answer:
b
Explanation:
A 10 N force is applied to a 2 kg mass, how fast will it be going in 10 sec?
Answer:
v = 50 ms^1
Explanation:
We know:
F = 10N
m = 2kg
t = 10 sec
F=ma
10 = 2*a
5=a
a = 5 ms^2
Now,
a=v/t
5 = v/10
50 = v
v = 50 ms^1
Calculate the speed of a 5 kg object that possesses 20 J of kinetic energy KE=1/2
mv2
2.83 m/s/s
2.83 m/s
2 m/s
4 m/s
Answer:
2.83 m/s
Explanation:
m = 5 kg,
KE = 20 J
KE = mv^2/2
20 = 5*v^2/2
5*v^2 = 40
v^2 = 8
v = [tex]\sqrt{8}[/tex] = 2.83 m/s
If a driver brings a car traveling at 22m/s to a full stop in 2.0 s with an acceleration of -8 m/s2 then how far did the car travel while braking?
Answer:
v=u + at
0= 22 + a*2
a = -22/2 = -11 m/s^2
The car's acceleration is -11 m/s^2
V^2 = u^2 + 2as
022^2 + 2*(-11)s
s = 22^2/22 = 22m
The car will travel 22 m before stopping.
Explanation:
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Answer this simple question and explain it briefly
Answer:
Explanation:
Here,
P=Pressure
V=Volume
Subscript meanings: 1 is before, 2 is after. So P1 is starting pressure, P2 is ending pressure.
In the P1V1 = P2V2 equation, P and V are inversely proportional.
It means that when a quantity will increase then another will decrease.
The pressure of a gas is directly proportional to the. The temperature when volume and amount of substance are constant. P1/T1 = P2/T2. Combined gas law: P1V1/T1 = P2V2/T2 Use the gas laws for pressure, volume, and temperature calculations. Avagadro's law and the Ideal gas law.
What is that force that simply pushes bodies through the planet's gravitational field
Answer:
Actually, it is not a force. It is a velocity attained by the bodies hence escaping through the gravitational field.
This velocity is called escape velocity.
[tex]{}[/tex]
Answer:
Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun. Weight is a measure of the force of gravity pulling down on an object. It depends on the object's mass and the acceleration due to gravity, which is 9.8 m/s2 on Earth. The gravitational force between two bodies is directly proportional to the product of the masses of those bodies and is inversely proportional to the square of the distance between them.
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-pound weight two-feet above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?
The number of times the athlete can lift the weight with single energy supplied by the Oreo cookie is 204 times and this number of times will increase if the efficiency increases.
Energy used by the athleteThe energy used by the athlete is calculated as follows;
E = Fd
where;
F is the applied force = 100 lb weight = 444.82 Nd is the distance = 2 ft = 0.61 mE = 444.82 x 0.61 = 271.34 J
Input energyThe Input energy of the athlete = 53 kcal = 221752 J
Number of times the athlete can lift the weight[tex]E = \frac{0utput \ energy}{1nput \ energy} \\\\0.25 = \frac{n(271.34)}{221752} \\\\271.34n = 55483\\\\n = \frac{55483}{271.34} \\\\n = 204[/tex]
Thus, the number of times the athlete can lift the weight with single energy supplied by the Oreo cookie is 204 times and this number of times will increase if the efficiency increases.
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1. A strobe pattern is made of marks on a piece of paper recorded every 0.1 s. Two adjacent marks near the middle of the pattern are 3.4 cm apart. Use this information to estimate the instantaneous speed at that point.
One mole of a monatomic gas is subjected to the following sequence of steps.
a. Starting at 300 K and 10 atm, the gas expands freely into a vacuum to double its volume.
b. The gas is next heated reversibly to 500 K at constant volume.
c. The gas is then reversibly expanded at constant temperature until its volume is doubled
d. The gas is finally reversibly cooled to 300 K at constant pressure
Required:
Calculate the values of q and w and the changes in U, H, and S.
Answer:
c
Explanation:
the gas is the reversible expanded and constant temperature until it's volume is doubled
A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. what is its acceleration?
Answer:
−2.373
Explanation:
Theta=14
ma=-mgsin(theta)
a= -gsin(theta)
a= -9.81sin(14)
a=−2.373
A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. -2.37 m/s² is its acceleration.
What is acceleration ?Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities, accelerations. The direction of the net force applied on the item determines the object's acceleration.
An object has positive acceleration if it is accelerating and travelling in the right direction. Positive acceleration was demonstrated in the first example by the speeding car. The acceleration is occurring in the same direction as the car's motion, which is forward and speeding up.
The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t). Using this, you may gauge the speed of velocity changes in meters.
θ = 14
Acceleration = -gsinθ
= - 9.8 × sin 14°
= -2.37 m/s²
Thus, A car moving 12.2 m/s starts to coast up a frictionless 14.0 degree hill. -2.37 m/s² is its acceleration.
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The volume
of this ice
would be
considered
what type of
property?
A. a chemical property
B. a rare property
C. a physical property
D. an undetermined property
Calculate change in height of a 2kg ball moving a speed of 10m/s up a frictionless ramp until it stops
The change in the height of the object is 5.1 m.
Conservation of mechanical energyThe principle of conservation of mechanical energy states that the total energy of an isolated system is always conserved.
The change in the height of the object is calculated by applying the principle of conservation of mechanical energy as follows;
P.E = K.E
[tex]mg \Delta h = \frac{1}{2} mv^2\\\\g \Delta h = \frac{1}{2}v^2\\\\\Delta h = \frac{v^2}{2g} \\\\\Delta h = \frac{(10)^2}{2(9.8)} \\\\\Delta h = 5.1 \ m[/tex]
Thus, the change in the height of the object is 5.1 m.
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A wooden supply raft is 2 m wide, 3 m long and 0.200 m deep. The raft and it’s occupants have mass of 700 kg. How deep will the sink below the water when floating?
58.3% of the raft will sink below the water when floating.
Volume of the raftThe volume of the raft is calculated as follows;
V = Lbh
V = 2 x 3 x 0.2
V = 1.2 m³
Density of the raftThe density of the raft is calculated as follows;
ρ = m/v
where;
m is the mass = 700 kgv is the volume = 1.2 m³[tex]\rho = \frac{700}{1.2} \\\\\rho = 583.33 \ kg/m^3[/tex]
Fraction of the raft below the water[tex]f = \frac{\rho\ _{raft}}{\rho \ _{water}} \times 100\%\\\\f = \frac{583.33}{1000} \times 100\%\\\\f = 58.3 \%[/tex]
Thus, 58.3% of the raft will sink below the water when floating.
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Physics Unit 5 CA - Conservation of
1. A student is pushing a car toward a gas station. They exert a constant force of 100 N
on the car for 40 m bring the car from rest to 5 m/s. What is the mass of the car?
Answer:
160 kg
Explanation:
f = ma
100 = m * a
a = v/t
v = distanc/time
dist = 40m
time = 8 sec
v = 40/8
a = 40/(64)= 5/8
m = 100*8/5= 160 kg
this is the question on the attachment
Explanation:
a) to allow for the metal tracks to expand on hot days.
reason:- as you have seen in summer the sunlight fall on these tracks and make the metal expand and fill that gaps.
hope this helps you
have a great day :)
how do you work this out
Answer:
A= v-u/t
= 8m/s - 18m/s ÷ 100s
= - 0.1m/s^2
Hope this helps!
If a yellow ball is rolled with an average speed of 5 m/s, how far will it travel
in 5 seconds? (just type the number, no need for units).
*
For the purpose of calculating the electric field strength by means of Gauss’s law, determine whether approximate cylindrical symmetry holds in each of the following situations.
Part (a) We have a 3.6-m long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (b) We have a 8.9-cm long copper rod of radius 1 cm, carrying a charge of 1.5 nC distributed uniformly along the rod’s length. We want to calculate the electric field strength at a point 4.9 cm from rod near its center.
TRUE FALSE
Part (c) A 1.9-m long wooden rod is glued end-to-end to a 1.9-m long plastic rod, both of radius 1 cm. The combined rod is then painted with an electrically charged paint so that it is covered with a uniform charge density, giving it total charge of 1.5 nC. We want to calculate the electric field strength at a point 4.9 cm from the rod near its center.
TRUE FALSE
Part (d) For the same charged rod as in part (c) we want to calculate the electric field strength at a point 8.9 m from each end of the rod.
TRUE FALSE
Answer:
1. True
2. False
3. True
4. False
Explanation:
a) In this case, length of the copper rod is 3.6 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.
b) In this case, length of the copper rod is 8.9 cm which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.
c) In this case, length of the copper rod is 3.8 m which is much larger than the distance 4.9 cm to the point at which electric field is to be determined. Therefore, yes, cylindrical symmetry holds.
d) In this case, length of the copper rod is 3.6 m which is of the same order of magnitudes the distance 4.9 cm to the point at which electric field is to be determined. Therefore, no, cylindrical symmetry does not hold.