Answer:
This question appear incomplete
Explanation:
This question appear incomplete because of the lack of options. However, a connected web of cells is generally referred to as the tissue. These cells are joined together by cell-cell adhesion (a form of bond) to form the tissue. The type of tissue being referred to here is the nervous tissue as this is the type of tissue present in the nervous system (including the central nervous system of which the brain is a part of).
can somone pls help asap???!!
The correct Answer: would be A B C E
Explanation:
Can i get brainlyest ?
Starting with a constant velocity of 45 km/h, a car accelerates for 35 seconds at an acceleration of 0.45 m/s2 . What is the velocity of the car at the end of the period of 35 seconds of acceleration?
Answer:
28.3 m/s
Explanation:
Vi = 45 Km/h = 12.5 m/s
Vf - Vi = at
Vf -12.5 = 0.45(35)
Vf= 28.3 m/s
(Need Help) (ASAP)
You are at the Grand canyon standing at the edge of a ledge 1,857 m high. you have a mass of 61 kg you decide to take a selfie to share with your science teacher when you get home witch causes you to wonder...
a. How much gravitational potential energy do you have standing on the edge of this cliff?
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, how fast would you be feeling?
Answer: Considering the gravitational field strength being 9.8...
A) 1,120,875J
B) 191.70m/s (2 DP)
Explanation:
explained in pic
You are at the Grand canyon standing at the edge of a ledge 1,857 m high. you have a mass of 61 kg you decide to take a selfie to share with your science teacher when you get home witch causes you to wonder,
a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.
What is Gravitational Potential Energy?
Gravitational potential energy is the potential energy with respect to the gravitational force. To possess gravitational potential energy, the object to be placed in a position in the gravitational field.
According to the question,
Gravitational potential energy = mgh
As given here, h = 1857 m
m = 61
Substituting the given formula,
Gravitational potential energy = 61 × 9.8 × 1875
=1,120,875 J.
Here, As Gravitation potential energy gained, the Kinetic energy lost.
So, Kinetic Energy K.E = 1,120,875 J.
As we know the formula,
K.E = 1/2 mv²
v² =K.E / (1/2 m)
= 1,120,875 × 2 / 61
v² =36,750
v = [tex]\sqrt{36750}[/tex]
= 191.702 m/s.
The speed was calculated as 191.702 m/s.
Thus,
a. The gravitational potential energy you have standing on the edge of this cliff is 1,120,875J.
b. If you were to trip and fall mid selfie, your stored potential energy would be converted to kinetic energy, as you are in motion period assume that all of the GPU calculated in part A would be converted to kinetic energy when you fall period if so, it was felt to be fast as 191.70m/s.
Learn more about Gravitational potential energy,
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1. A ball is thrown straight up.if the launch velocity is 15 m/s, at what velocity will the ball return to the thrower's hand?
2. A boat moves across a river going 18 m/s. At the same time there is a current flowing at a right angle to the boat at 6 m/s. What is the resulting velocity of the boat?
Answer:
1) The velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
Explanation:
1) Let suppose that ball experiments a free fall, that is an uniform accelerated motion, in which effects from gravity and Earth's rotation can be neglected. The velocity of the ball is represented by the following equations of motion:
Position
[tex]v_{o}\cdot t -\frac{1}{2}\cdot g\cdot t^{2} = 0[/tex]
[tex]t\cdot \left(v_{o}-\frac{1}{2}\cdot g\cdot t \right) = 0[/tex] (1)
Velocity
[tex]v = v_{o}-g\cdot t[/tex] (2)
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{o}[/tex] - Initial velocity of the ball, measured in meters per second.
[tex]v[/tex] - Final velocity of the ball, measured in meters per second.
From (1), we get the time when the ball returns to the thrower's hand:
[tex]v_{o}-\frac{1}{2}\cdot g\cdot t = 0[/tex]
[tex]t = \frac{2\cdot v_{o}}{g}[/tex]
And then we apply this result in (2):
[tex]v = v_{o}-g\cdot \left(\frac{2\cdot v_{o}}{g} \right)[/tex]
[tex]v = -v_{o}[/tex] (3)
Then, the velocity of the ball return to the thrower's hand is -15 meters per second.
2) The resulting velocity of the boat ([tex]\vec v_{B}[/tex]) is represented by the vectorial sum of the velocity of the boat relative to the river ([tex]\vec v_{B/R}[/tex]) and the velocity of the river ([tex]\vec v_{R}[/tex]), both measured in meters per second, that is:
[tex]\vec v_{B} = \vec v_{R}+\vec {v}_{B/R}[/tex] (4)
If we know that [tex]\vec v_{R} = 6\,\hat{i}\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{B/R} = 18\,\hat{j}\,\left[\frac{m}{s} \right][/tex], then the resulting velocity of the boat is:
[tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex]
The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].
There is an attachment. There is an attachment on here.
Answer:
5 i think
Explanation:
You are working as an intern for a meteorological laboratory. You are out in the field taking measurements from a balloon that is carrying equipment designed to measure electric fields in the atmosphere. Your supervisor has asked you to determine the average volume charge density at a certain height in the air. When the balloon is at an altitude of 500 m above the ground, the electric field is measured to be 160 N/C directed downward. At 800 m above the ground, the electric field is 120 N/C downward.
(a) Determine the average volume charge density (in C/m3) in the layer of air between these two elevations. (Enter the magnitude.)
(b) Is this layer of air positively or negatively charged?
positively charged
negatively charged
Answer:
a. -6.99 × 10³⁰ C/m³ b. The layer of air is negatively charged.
Explanation:
With E₁ = electric field at 500 m above the ground = 160 N/C (it is negative since it is directed downwards).
Also, with E₂ = electric field at 800 m above the ground = 120 N/C (it is negative since it is directed downwards).
The total flux, Ψ = ∫E.dA = E₁dAcosθ + E₂dAcosθ
For the 800 m surface E is parallel to dA, that is = 0° and For the 500 m surface E is anti-parallel to dA, that is = 180°
Ψ = ∫E₁dAcos180° + ∫E₂dAcos0°
= -∫E₁dA + ∫E₂dA
= -E₁∫dA + E₂∫dA
= -E₁4πR² + E₂4πR²
= (-E₁ + E₂)4πR² where R = radius of earth = 6.4 × 10⁶ m
= (-160 N/C + 120 N/C)4π(6.4 × 10⁶ m)²
= - 40 N/C)4π(6.4 × 10⁶ m)²
= -20588.74 × 10¹² C
= -2.058874 × 10¹⁶ C
≅ -2.06 × 10¹⁶ Nm²/C.
The since charge, Q = Ψ/ε₀, the total charge through the area is thus
Q = Ψ/ε₀
= -2.06 × 10¹⁶ Nm²/C ÷ 8.854 × 10⁻¹² F/m
= -0.233 × 10²⁸ C/m²
= -2.33 × 10²⁸ C/m².
So, the charge in the volume = charge net charge of surface × width of volume. So the charge in the volume Q' = QΔh = Q(h₂ - h₁) where h₁ = 500 m and h₂ = 800 m
Q' = Q(h₂ - h₁)
= -2.33 × 10²⁸ C/m²(800 m - 500 m)
= -2.33 × 10²⁸ C/m²(300 m)
= -699 × 10²⁸ C/m³
= -6.99 × 10³⁰ C/m³
b. Since Q' = -6.99 × 10³⁰ C/m³, the layer of air is negatively charged.
A piece of iron of mass 200g and tempreture 300°C is dropped into 1.00 kg of water of tempreture 20°C. Predict the final equilibrium of the water.
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.
[tex]Q_{iron} = Q_{water}[/tex]
Heat transfer can be found by means of the following equation.
[tex]Q_{iron}=m*C_{piron}*(T_{i}-T_{f})[/tex]
where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]
[tex]Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})[/tex]
Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
[tex]0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C][/tex]
During a laboratory experiment, the temperature of the gas in a balloon is varied and the
volume of the balloon is measured.
A)
Which quantity is the independent variable?
B)
Which quantity is the dependent variable?
Which force results from charged particles
Answer:
electromagnetic force
Explanation:
A toy train engine is rolling freely at a constant speed on a level piece of track. The train engine
collides with a stationary truck, and joins with it. Before the collision the train engine is travelling at
0.30 m/s, and has a mass of 700g. If the stationary truck has as mass of 400g. calculate the speed
of the joined engine and truck immediately after the collision.
(PLEASE answer with working).
Answer:
v₃ = 0.19 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of the linear momentum, which is defined as the momentum before the collision is equal to the momentum after the collision.
ΣPbefore = ΣPafter
[tex]P=m*v[/tex]
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
Let's imagine that the toy train moves to the right, this movement is taken as positive.
[tex](m_{1}*v_{1})+(m_{2}*v_{2})= (m_{1}+m_{2})*v_{3}[/tex]
where:
m₁ = mass = 700 [g] = 0.7 [kg]
v₁ = velocity = 0.3 [m/s]
m2 = mass = 400 [g] = 0.4 [kg]
v₂ = 0 (stationary truck, there is no movement)
v₃ = final velocity of the joined engine and truck.
Now replacing:
[tex](0.7*0.3)+(0.4*0)= (0.7+0.4)*v_{3}\\0.21 = 1.1*v_{3}\\v_{3}= 0.19 [m/s][/tex]
wassup people of this wor;d
An applied force is a force that is applied to an object by a person or another object.
True or False
Answer:
true
Explanation:
What is the normal force acting on the 7.50 kg box shown?
The answer would be 31 N. This is because Fn = 7.5*9.8*sin(25).
What reaction/soccur between an acid and base.
Answer:
neutralization reaction
Explanation:
The reaction of an acid with a base is called a neutralization reaction.
es una reacción química que ocurre entre un ácido y una base produciendo sal y agua. ... Se mezcla un ácido fuerte con una base fuerte: Cuando esto sucede, la especie que quedará en disolución será la que esté en mayor cantidad respecto de la otra.
A kangaroo hopped at 22 meters per second, saw a skunk and then
hopped at 52 meters per second. This is an example of which of the
following?
Acceleration
Velocity
Speed
motion graphs
Answer:
this is an example of acceleration
The ball in this activity could reach much greater speeds if not for the loss of energy in many different transformations. There is one major energy loss that is causing the ball to move more slowly, what is that energy? What force is acting upon the ball to create this loss of energy?
Please help quick!
Answer:
friction acts upon the ball when it is sliding down the slide and this action creates thermal energy
At which temperature could air hold the most water vapor?
O A. 24°C (75°F)
B. 2°C (36°F)
C. 35°C (95°F)
O D. 13°C (55°F)
SUBMIT
Explanation:
35 maybe hope it's right
What is the life cycle of our sun?
Answer:
the sun is currently a mian sequence star
and will remain so for another 4-5 billion
Helppppppppppp I’ll give brainliest
Answer:
[tex] \boxed{(i).... 40cm}\\ \boxed{(ii).... 18cm}\\ \boxed{(iii).... 7.5N
}\\ [/tex]
Explanation:
[tex](i).... \\ the \: weight \: of \: the \: rod \: acts \: at \: the \: center \\ which \: is \: the \: half \: of \: 80cm = 40cm \\ (ii).... \\ let \: the \: distance \: be \: \boxed{d}...then \: 22 + d = 40 \\ d = 40 - 22 = 18cm \\ (iii).... \\let \: the \: weight \: of \: the \: metal \: rod \: be \: \boxed{ w} \\ taking \: the \: moment \: about \: the \: string \to \\ 15 \times (22 - 13) = 18w \\ 18w = (9)(15) \\ 2w = 15 \\ w = 7.5N[/tex]
♨Rage♨
15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters
east, 40 meters west, and finally 50 meters east. What is her total distance
and displacement, respectively?
Answer:
Long question good luck:)..............
Explanation:
State the career function of mining engineering
Answer:Mining engineers are responsible for the effective, safe and profitable operation of mining undertakings. They are mining experts and engineers and have a background in geology as well as civil, mechanical and electrical engineering.
Explanation:
What is a Super Massive Black Hole?
Answer:
I dont really know much but i know that it swallow anything it comes across in space.
May someone please help?
Air is cooled and
becomes
[denser/less dense).
Less dense air
[rises/sinks).
Denser air
(rises/sinks).
The process repeats, producing a
Air is heated and
becomes
[denser/less dense]
within the gas
Air is cooled and
becomes denser.
Denser air sinks.
The process repeats, producing a (i wasn't sure about this one)
Air is heated and becomes less dense.
Less dense air rises.
hope this helps!
The pressure of liquid varies as per
its depth
Answer:
Yes.
Explanation:
The pressure varies as per the depth of the container
SILVER PACKAGE
The work done in lifting a 10 kg load of bricks to the top of a building 20 m high is (g = 9.8 ms-2
a) 98
b) 200 J
c) 980
d) 1960)
The radius of Mars is 3397 km; and its mass is 6.42x10²³ kg. What is the acceleration due to gravity at an altitude of 485 km above the surface of Mars?
Answer:nfeergnjknrtgrth
Explanation:
Answer:
F = GMm/r^2
MA = GMm/r^2
A = GM/r^2
A = (6.67 * 10^-11)(6.42 * 10^23) / (3.37 * 10^6)^2
A = 3.77 m/s^2
Explanation:
Why is my life like this
Answer:
stop it get some help
:)
Explanation:
Answer:
My life be like oooh ahhh ooh ah
Read the elapsed time on the stopwatch and answer the questions.
In which digit is there the least amount of confidence?
How many significant figures does this measurement have? I need help quick
Answer:
1.7
2.4
Explanation:
Answer:
1.7
2.4
Explanation:
What made "Lunar Orbit Rendezvous" such an attractive method to get to the Moon that it was selected by NASA in July, 1962
Answer:
I don't know the answer
Can two objects be at the same distance from a single point but be in different positions? Why or why not?
Plsssss help meee
Answer:Position is an object's location in relation to a reference point. Yes. Both objects are one unit away (the same distance), but they occupy different locations in space because they are in different directions.
Explanation: Hope this helps!!
Yes it is possible for two objects be at the same distance from a single point but be in different positions.
We have two objects at the same distance from a single point but are in different positions.
We have to determine whether this statement is true or not.
What is a Reference frame?A reference frame can be considered as the coordinate frame with respect to origin of which the physical quantities such as velocity, displacement , acceleration etc. of any other object in different reference frame is measured.
According to the question -
YES, Two objects can have same distance from a single point or a reference point (say origin) and can be at different positions. For example - Two different points with Position vectors -
P(1) = i + j
P(2) = - i + j
are located. Now, distance of point P(1) from origin will be - [tex]\sqrt{1+1}=\sqrt{2}[/tex] and that of point P(2) from origin will be - [tex]\sqrt{(-1)^{2} +(1)^{2} } =\sqrt{2}[/tex]
Hence, yes it is possible for two objects be at the same distance from a single point but be in different positions.
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