Answer:
E = V / d
Explanation:
In a charged capacitor an electric field is established that goes from the positive to the negative plate, this field is constant,
the potential difference is
D = E d
in this case they do not give the difference in potential V and the distance between the plates d
E = V / d
You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Object 3 is a sphere of copper.
a. the density of tin is 5.75g/cm2. What is the mass of object 1 in kg if the rectangular block has a volume of 1.34L?
b. what is the volume in cubic inches of object 2 if the cube of aluminum 7.58 inches on a side?
c. what is the mass in kg of object 2? the density of aluminum is 2.70g/cm3
d. what is the volume in cm3 of object 3 if the sphere of copper has a diameter 8.62cm? the volume of the sphere is 4 {pi}^3/3
e. what is the mass in kg of object 3? Copper has a density of 8.96g/cm3
Answer: a. m = 7.7 kg
b. V = 435.52 in³
c. m = 1927 kg
d. V = 335.37 cm³
e. m = 3 kg
Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.
The formula for density is
[tex]\rho=\frac{m}{V}[/tex]
Density's unit in SI is kg/m³, but it can assume lots of other units.
Some unit transformations necessary for the resolution of the question:
1 L = 1 dm³ = 1000 cm³
1 in³ = 16.3871 cm³
1 g = 0.001 kg
a. V = 1.34 L = 1340 cm³
[tex]\rho=\frac{m}{V}[/tex]
[tex]m=\rho.V[/tex]
m = 5.75 * 1340
m = 7705 g => 7.705 kg
Mass of object 1 with volume 1.34L is 7.7 kg.
b. A cube's volume is calculated as V = side³
V = 7.58³
V = 435.52 in³
Volume of object 2 is 435.52 in³.
c. Using 1 in³ = 16.3871 cm³ to change units:
V = 435.52 * 16.3871
V = 713689.4 cm³
Then, mass will be
[tex]m=\rho.V[/tex]
m = 2.7 * 713689.4
m = 1926961.4 g => 1927 kg
Mass of object 2 is 1927 kg.
d. Volume of a sphere is calculated as [tex]V=\frac{4}{3}.\pi.r^{3}[/tex]
Diameter is twice the radius, then r = 4.31 cm.
Volume is
[tex]V=\frac{4}{3}.\pi.(4.31)^{3}[/tex]
V = 335.37 cm³
Volume of object 3 is 335.37 cm³.
e. [tex]m=\rho.V[/tex]
m = 8.96 * 335.37
m = 3004.91 g => 3 kg
Mass of object 3 is 3 kg.
In a crash test, a car with a mass of 1600 kg is initially moving at a speed of 20 m/s just before it collides with a barrier. The final speed of the car after the collision is zero. The original length of the car is 4.50 m , but after the collision, the smashed car is only 3.60 m long.
Required:
a. What is the average speed Of the car during the period from first contact with the barrier to the moment the car comes to a stop? You may assume the force that the barrier exerts on the car is constant during this period.
b. How much time elapses between the moment the car makes first contact with the barrier and the moment it comes to a stop?
c. Making the very rough approximation that the large force that the barrier exerts on the car is approximately constant during contact, determine the approximate magnitude of this force?
Answer:
The answer to the given points can be defined as follows:
Explanation:
In point 1:
[tex]\bold{v_f^2= v_i^2+2as}\\\\\to v_f=0\\\\\to v_i=20 \frac{m}{s}\\\\\to s= 4.50\ m -3.60 \ m \\\\[/tex]
[tex]=0.9 \ m \\[/tex]
put the value in the above formula:
[tex]\to 0= 20^2+2 \times a \times 0.9\\\\\to -1.8\ a=400\\\\\to -a= \frac{400}{1.8} \\\\ \to a= -222.22\ \ \frac{m}{s^2}[/tex]
[tex]\bold{v_f=v_i+at}\\\\\to 0=20+ (-222.22)t\\\\\to 222.22t=20\\\\\to t=\frac{20}{222.22}\\\\\to t= 0.0900 \ s\\\\\to v_{avg}=\frac{s}{t}=\frac{0.9}{t}= 10\ \frac{m}{s}[/tex]
for point 2:
[tex]t= 0.0900 \ s -\text{found above}[/tex]
for point 3:
[tex]\to |a| = 222.22 \frac{m}{s^2} \text{found above}\\\\\to \bold{|F| = m \cdot |a|}\\\\[/tex]
[tex]=1600 \ kg \times 222.22 \ \frac{m}{s^2} \\\\= 3.55\times 10^{5} \ N[/tex]
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water
Answer:
See explanation below
Explanation:
This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.
First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:
V = V₀ + gt (1)
X = V₀t + gt²/2 (2)
In this case, g = 9.8 m/s²
Now, let's see the displacement and velocity for each given time:
a) For t = 0.5 s
V = 14 + (9.8)*0.5
V = 18.9 m/s
X = (14*0.5) + (9.8)(0.5)²/2
X = 7 + 1.225
X = 8.225 m
b) For t = 1.00 s
V = 14 + (9.8)*1
V = 23.8 m/s
X = (14*1) + (9.8)(1)²/2
X = 14 + 4.9
X = 18.9 m
c) For t = 1.5 s
V = 14 + (9.8)*1.5
V = 28.7 m/s
X = (14*1.5) + (9.8)(1.5)²/2
X = 21 + 11.025
X = 32.025 m
d) For t = 2 s
V = 14 + (9.8)*2
V = 33.6 m/s
X = (14*2) + (9.8)(2)²/2
X = 28 + 19.6
X = 47.6 m
e) For t = 2.50 s
V = 14 + (9.8)*2.5
V = 38.5 m/s
X = (14*2.5) + (9.8)(2.5)²/2
X = 35 + 30.625
X = 65.625 m
Hope this helps
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
WILL MARK BRAINLIEST IF CORRECT. Worth Lots Of Points. NEEDS TO BE WITH EXPLANATION PLEASE. PLEASE HELP!
A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resistance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter
Answer:
Explanation:
For resistance , the expression is as follows .
R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .
cross sectional area = π x ( .5 x 10⁻³ )²
S = .785 x 10⁻⁶ m²
Putting the values
R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶
= 1.796 x 10⁻² ohm .
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
Why do people eat bo oty
Answer: I don't know my dude
Explanation:
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
Which of the following is not a true statement?
A
B
C
D
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
0.2 mph
4.8 mph
5.5 mph
144.1 mph
it is actually science i couldn't find the word science
Answer:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
Answer:
38.7 mph
Explanation:
I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.
77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m
Answer:
a
Explanation:
i just took the test
a bus initially at rest accelerated of 4m/S2 at the end of 10 second find average velocity
Answer:
V = 20m/s
Explanation:
Given the following data;
Acceleration = 4m/s²
Time = 10 secs
Initial velocity = 0m/s (since it's at rest).
To find the average velocity, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
V = 0 + 4*10
V = 40m/s
To find the average velocity;
Average velocity = (U + V)/2
Average velocity = (0 + 40)/2
Average velocity = 40/2
Average velocity = 20m/s
Most new jobs in the United States will be in the _____.
in the service producing sector
First to answer gets brainliest
A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop
Answer:
the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
Explanation:
Given that;
mass of the drop m = 8.0×10⁻¹⁵ kg
distance d = 16 cm = 0.16 m
potential difference between the plates V = 2.44 kV = 2440 v
acceleration of gravity g = 9.81 m/s²
the magnitude of the charge on the drop = ?
weight is balanced by the electrostatic force
weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴
we know that; V = Ed
E = V/d = 2440 / 0.16 = 15200 v/m
Electrostatic force = qE
so weight = qE
q = weight / E
q = 7.848 × 10⁻¹⁴ / 15200
q = 5.163 × 10⁻¹⁸ C
Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
I NEED HELP WITH THIS QUESTION ASAP PLEASE!
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
What is the value of the angle of inclination of the slide?
Answer:
63°
that's my answer
but then I am sorry if I'm wrong
Explanation:
90-27 = 63°
A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such devices because they can simulate the same Accelerations that are experienced in certain types of flight. At which angular velocit would a space station of 30 m radius have to rotate to simulate Earth graivty?
Answer:
ω = 0.571 rad/s
Explanation:
given data
radius = 30 m
solution
we take here g = 9.8 m/s²
and g is express as
g = r × ω² ....................1
put here value and we get
9.8 = 30 × ω²
solve it we get
ω = 0.571 rad/s
) Explain why the foil is attracted at first by the charged rod. Consider any charge that exists in the uncharged foil. (You will consider later interaction between the rod and foil in the last question. Just think about the first interaction when answering this question.) [
Answer:
The attraction is due to the induced charge.
Explanation:
When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.
This induced load occurs if importing the plate load
The attraction is due to the induced charge.
When you jump, you push down on the earth and it pushes back up against you. The earth pushing up against you is what causes you to go into the air. Why doesn’t your push cause the earth to go down if your push on the earth is equal and opposite of the earth's push on you?
That's a great question !
The answer is: It does !
A push on an object causes the object to accelerate in the direction of the force.
The less mass the object has, the more the force accelerates it.
Now, when you jump, the forces on you and on the Earth are equal forces.
The up force on you causes you to accelerate up by some amount.
The down force on the Earth causes the Earth to accelerate down by some amount.
The Earth's mass is something like 5,972,000,000,000,000,000,000,000 kg, while your mass is something like 50 kg.
The Earth has something like 119,400,000,000,000,000,000,000 times as much mass as you have.
So your acceleration is something like 119,400,000,000,000,000,000,000 times as great as the Earth's acceleration.
==> The Earth's downward acceleration, caused by your jump, is there. It's just too small to notice.
BUT . . . That's the reason why seismometers (instruments to detect and measure the vibrations from distant earthquakes) have to be located as far as possible from cities and busy roads.
In places that are too close to cities and roads, the Earth's surface is always vibrating, wiggling, jiggling, heaving and weaving, in reaction to the forces of people walking around, cars and trucks driving around, even rain falling down. And kids jumping up and down !
In such places, these people-motions are louder and stronger than the vibrations coming from distant earthquakes. Seismometers wouldn't work there.
A hockey puck with mass 0.30 kg is sliding along the ice with initial speed of 12.68 m/s. A hockey player is heading toward the puck with his stick in hand. After the player strikes the puck, the puck reverses its direction and is traveling at double its speed before the strike. If the collision occurs in 0.05 s, what is the magnitude of the force the hockey player's stick applied to the puck
Answer:
F = 228.24 N
Explanation:
According Newton's 2nd Law, the impulse on one object is equal to the change in momentum of that object.I = F*Δt = Δp = pf - po (1)where pf = final momentum = m*vf
p₀ = initial momentum = m*v₀
Since after the strike, the puck reverses its direction and travels at double its speed before the strike, that means that vf = -2*v₀.Replacing in the right side of (1), we have:[tex]m*v_{f} - m*v_{o} = -2*v_{o} -m*v_{o} = -3*m*v_{o} = -3*0.3kg*12.68m/s = -11.41m/s (2)[/tex]
Replacing Δt = 0.05s, and solving for F in (1):[tex]F_{net} = \frac{-11.41m/s}{0.05s} = -228.24 N (3)[/tex]
which means that the force is applied in a direction opposite to the initial velocity of the puck.The magnitude of the force is just 228.24 N.Help me I don't know what I'm doing
Answer:
C the metal handle because it is a good conductor
Answer:
D.
Explanation:
Although the metal handle will last longer, if heated up enough it could burn her hand.
Solve this chemical equation: CH3CH2OH+__O2=CO2+__H2O
Answer:
some kind of chemical of which i do not know
Explanation:
Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws a rock straight upwards with an initial speed of 15 m/s, while the other person throws their rock straight downwards with an initial speed of 15 m/s. If both of the rocks miss the building and continue down to the street,
a. How long will it take for the rock thrown upwards to reach the ground?
b. How long will it take for the rock thrown downwards to reach the ground?
c. How fast will the upward thrown rock be travelling just before it hits the ground? The downward thrown rock?
Answer:
C
Explanation:
Hope this helps!!!!
anteaters are animals with long tongues that they used to wear a ant they sometimes free by poking their tongues into a bed and licking the ants up if an anteater is born with shortened tongue this would be an example of a
Answer: an ant eater who can’t eat
Explanation:
Astronaut 1 has a mass of 70 kgkg. Astronaut 2 has a mass of 80 kgkg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in NN). Astro 1 visits a planet with gravitational acceleration 7.0 m/s2m/s2. What must be Astro 2 planet's agag to equal Astro 1's weight
Answer:
g₂ = 39 m / s²
Explanation:
Mass is the amount of matter that a body has, so it does not change, but the weight is the force with which the planet attracts this matter, this value does change
W = m g
let's use subscript 1 for the first planet and subscript 2 for the second planet
W₁ = m₁ g₁
W₂ = m₂ g₂
They give us the values of the masses and the value of g₁ = 7.0 m / s, let's find the value of the acceleration of gravity on Planet 2
They tell us that the weight of the two astronauts is the same therefore
W₁ = W₂ = W
let's match the expressions
m₁ g₁ =m₂ g₂
g₂ = [tex]\frac{m_1}{m_2} \ g_1[/tex]
let's calculate
g₂ = [tex]\frac{70}{80} \ 7.0[/tex]
g₂ = 39.2 m / s²
g₂ = 39 m / s²