Answer:
sorry don't know i just need points
Explanation:
What is also known as watered carbons
Answer:
The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.
ASAP it’s timed!!
An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s? (Formula: Fema)
0.89 N
1.2 N
11 N
30 N
Answer:
30N is the correct answer
What are four pieces of evidence for continental drift?
How many seconds are there in a calendar year,given that a calendar year has 365.24 days in it?
Answer:
31556736
Explanation:
Since there are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365.24 days in a year, then you multiply all of those.
60*60*24*365.24 =
31556736
What are indicative acts? FORENSICS
Answer:
Actions or behaviors that may have taken place before death.
Do you wanna know how I know? No.... You do not wanna know how I know that you know what I known that you know I know
Explanation:
You Know?
if 56 j of work energy is required to lift a dumbbell to a height of 7m, what is the weight of the dumbbell?
Answer:
8 N
Explanation:
work=force*distance
force=work/distance
=56/7
=8
The graph shows the speeds of two bicyclists on their way to school. what is the speed of bicycle B
Answer: 0.3
Explanation: Jusy took it and got a 100
Answer: 0.3
Explanation: This is right trust me I answered this and it was correct!!
calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s
Answer:
height = 45 m
Explanation:
Initial velocity = 0 m/s
Final velocity = 30 m/s
Acceleration due to gravity = +10m/s^2 ( because ball is going down )
Using the eqn of motion
[tex] {v}^{2} - {u}^{2} = 2gh[/tex]
where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height
So,
[tex] {30}^{2} - {0}^{2} = 2 \times 10 \times h[/tex]
[tex] = > 20h = 900[/tex]
[tex] = > h = \frac{900}{20} = 45[/tex]
answer them all please!!
Answer:
1. stomach
2. arteries
3. trachea
4. spinal cord
Which two factors affect the amount of force seen in objects?
Answer:thanks
Explanation:xixo
gv
amp, is named after a person who was the person
Answer:
Explanation:
Named for 19th-century French physicist André-Marie Ampère, it represents a flow of one coulomb of electricity per second.
which force is a contact force
what is marketing according to AMA
Answer:
The new 2007 AMA defini- tion is as follows: Marketing is the activity, set of institutions, and processes for creating, communicating, delivering, and exchanging offerings that have value for customers, clients, partners, and society at large.
Explanation:
Man has never directly reached the inner layers of the Earth. But what phenomena allow to determine its chemical composition?
A)Surface wave propagation.
B)Plate displacement
C)Emanation of lava in volcanoes
D)Changes in the Earth's landscape
E)Waves formed in tidal waves
What is the modern periodic table of elements?
A. A chart organizing all elements by their atomic mass
B. A list of all elements occuring in nature
C. A chart organizing all elements by their atomic number
D. A list of all elements created by humans
A transformer has 1000 turns on the primary coil.
The voltage applied to the primary coil is 230V a.c,
How many turns are on the secondary coil if the output
voltage is 46 V a.c.?
A)20 B)200 C)2000 D4000 E)8000
Answer:
200
Explanation:
Given parameters:
Number of turns in primary coil = 1000 turns
Primary voltage = 230V
Secondary voltage = 46V
Unknown:
Number of turns in secondary coil = ?
Solution:
To solve this problem, we have to use the expression for solving problems related with transformers;
[tex]\frac{Vs}{Ns}[/tex] = [tex]\frac{Vp}{Np}[/tex]
Vs is the secondary voltage
Ns is the number of secondary turns
Vp is the primary voltage
Np is the number of primary turns
Now insert the parameters and solve;
[tex]\frac{46}{Ns}[/tex] = [tex]\frac{230}{1000}[/tex]
46 x 1000 = 230Ns
Ns = 200
1. A plane flying horizontally at an altitude of 490 meters and having a velocity of 250 meters per second east, drops a supply packet to a work crew on the ground. It falls freely without a parachute. [Assume no wind and negligible air resistance.] A. Calculate the time required for the packet to hit the ground. (25 pts)
Answer:
t = 10 s
Explanation:
Given :
Altitude, h = 490 m
Initial velocity of a plane, u = 250 m/s
To find :
The time required for the packet to hit the ground. When it hits the ground its velocity will be 0. Let t is the required time. It can be calculated as follows :
[tex]t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 490}{9.8}} \\\\t=10\ s[/tex]
So, it will take 10 s to reach the ground.
The gradual temperature change from one season to the next is due to what? view attachment
A projectile is launched horizontally at a speed of 40 meters per second from a platform located a vertical distance h above the ground. The projectile strikes the ground after time at horizontal distance from the base of the platform. [Neglect friction.]
Calculate the vertical distance, h, if the projectile's total time of flight is 4.5 seconds. You must show work to earn full credit with the minimum 3 steps:
1. Equation used
2. Substitution with correct units
3. Final answer with correct units
Answer:1
Explanation:t=rad2h/g
The launch of projectiles allows to find the height of the body launched horizontally is:
The height is 99.2 m
Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is the gravity acceleration.
In the attachment we see a diagram of the movement, as the body is thrown horizontally, its initial vertical speed is zero.
y = y₀ + [tex]v_o_y[/tex] t - ½ g t²
Where y, y₀i are the current and initial heights, respectively, [tex]v_{oy}[/tex] the vertical initial velocity, g the acceleration of gravity and t the time.
0 =y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 4.5²
y₀ = 99.2 m
y₀ = h = 99.2 m
This is the initial height of the object when it is thrown.
In conclusion using the projectile launch we can find the height of the horizontally launched body is:
The height is 99.2 m
Learn more here: brainly.com/question/10903823
What is the acceleration of an object with a mass of 367 kg when a net force of 183 N is applied to it
Answer:
[tex]0.5m/s^{2}[/tex]
Explanation:
F= m x a
a= [tex]\frac{F}{m}[/tex] = [tex]\frac{183 N}{367 kg}[/tex] = [tex]\frac{0.4986 N}{kg}[/tex] = [tex]\frac{0.4986 kg*m/s^{2} }{kg}[/tex] = [tex]0.5 m/s^{2}[/tex]
Hopefully, this is right!
Which of the statements are true about the collision of two objects with different
masses based on Newton's laws of motion?
A. Each object experiences an equal and opposite force.
B. The more massive object will exert a greater force on the less massive object.
Each object will accelerate in an opposite direction from the other but at a
different rate.
D. The less massive object will accelerate due to the force of the collision, but the
more massive object will not.
Answer:
A. TRUE
B. TRUE
C. It depends on the mass of both the objects
D. TRUE
A car turns from a road into a parking lot and into an available parking space. The car's initial velocity is 4 m/s [E 45° N]. The car's velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.
Answer:
Explanation:
From the given information:
The car's initial velocity = 4 m/s in the direction of east 45° due north
We can therefore express the vector of this component form as:
v₁ = (4 m/s) (cos(45º)i + sin(45º)j)
v₁ = (2.83 m/s)i + (2.83 m/s)j
Similarly, the car's final velocity = 4 m/s in the direction of the east side 10º north
∴
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ = (3.94 m/s) i + (0.695 m/s) j
From the first equation of motion
v = u + at
Making acceleration "a" the subject of the formula, we have:
a = (v - u )/t
a = (v₂ - v₁)/t
a = (0.370 m/s²) + (-0.711 m/s²)
The magnitude of the avg. acceleration is:
[tex]|| a||= \sqrt{(0.370 m/s^2)^2 + (-0.711 m/s^2)^2)[/tex]
[tex]|| a||= 0.8015 \ m/s^2[/tex]
And;
The direction can be determined by taking the tangent of the acceleration:
i.e.
[tex]tan(\theta) = \dfrac{-0.711 m/s^2}{ 0.370 m/s^2}[/tex]
[tex]tan(\theta) = -1.9216[/tex]
[tex]\theta = tan^{-1} ( -1.9216 )[/tex]
[tex]\mathbf{\theta = -62.51 ^0}[/tex]
Thus, the direction of the angle is approximately S 62.51º E
among 1 litre boiling water and red hot glowing nail which one has more amount of heat and why
Answer: The Water
Explanation:
The boiling litre of water has more volume and so has more heated molecules per cubic metre so it will have a higher amount of heat than the glowing nail.
To practically test this, put the glowing nail into a litre of cold water and record the final temperature when the nail has cooled down. Repeat this with the boiling litre of water. You will notice a higher final temperature with the boiling water indicating that it had more heat.
An 85kg mass accelerates at 5.5m/s2 what force was applied
Answer:
467.5 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 = 85 × 5.5
We have the final answer as
467.5 NHope this helps you
compare the 3 different types of waves
Answer:Mechanical waves, electromagnetic waves, and matter waves.
Explanation: does are the waves
In consideration of the above diagram, a projectile is launched at 45 with an initial horizontal velocity (V0x) of 3 m/s. Calculate the Vox which is the actual launch velocity at 45 in m/s.
CFISD too huh? :D
also tryna get the answer
please help .-. its for physics
Answer:
your answer is correct
How does a jack spread out the work over a large distance?
calculate the velocity of an object moved around a circle with a radius of 1.65m and an acceleration of 3.5 m/s^2
Answer:
2.4m/s
Explanation:
A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How FAST (m/s) is the car moving after this time?