which of the following to all food chains depend on in an ecosystem
Answer:
The sun is the ultimate source of energy for all food chains. Through the process of photosynthesis, plants use light energy from the sun to make food energy. Energy flows, or is transferred through the system as one organism consumes another.
A child pulls a wagon across the grass so that it accelerates using a force of 50 N at an angle of 42 degrees above the ground. The loaded wagon has a mass of 12 kg. If the coefficient of friction between the wagon and grass is 0.64. What is the acceleration of the wagon? Describe the motion of the wagon.
Answer:
[tex]-1.398\ \text{m/s}^2[/tex]
Decelerating or slowing down
Explanation:
F = Force = 50 N
[tex]\theta[/tex] = Angle force is being applied = [tex]42^{\circ}[/tex]
[tex]\mu[/tex] = Coefficient of friction = 0.64
m = Mass of wagon = 12 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Normal force is given by
[tex]N=mg-F\sin\theta[/tex]
Frictional force is given by
[tex]f=\mu N\\\Rightarrow f=\mu (mg-F\sin\theta)[/tex]
The force balance is given by
[tex]F\cos\theta-f=ma\\\Rightarrow \dfrac{F\cos\theta-\mu (mg-F\sin\theta)}{m}=a\\\Rightarrow a=\dfrac{50\times \cos42^{\circ}-0.64(12\times 9.81-50\times\sin42^{\circ})}{12}\\\Rightarrow a=-1.398\ \text{m/s}^2[/tex]
The acceleration of the wagon is [tex]-1.398\ \text{m/s}^2[/tex]. The negative sign indicates that the wagon is decelerating or slowing down.
The acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
Given data:
The magnitude of pulling force is, F = 50 N.
The angle of inclination is, [tex]\theta = 42^{\circ}[/tex].
The mass of wagon wheel is, m = 12 kg.
Coefficient of friction between wagon and grass is, [tex]\mu =0.64[/tex].
The given problem is based on the concept of frictional force. The standard expression for the frictional force is,
[tex]f= \mu \times N[/tex]
Here, N is the normal force and its value is,
[tex]N=mg-Fsin \theta[/tex]
And the net force acting on wagon is,
[tex]F' = Fcos\theta -f\\\\ma = Fcos\theta -(\mu(mg-Fsin \theta))\\\\a = \dfrac{Fcos\theta -(\mu(mg-Fsin \theta))}{m}[/tex]
Here, a is the acceleration of wagon.
Solving as,
[tex]a = \dfrac{50 \times cos42 -(0.64(12 \times 9.8-(50 \times sin42)))}{12}\\\\a=-1.398 \;\rm m/s^{2}[/tex]
Thus, we can conclude that the acceleration of the wagon is [tex]-1.398 \;\rm m/s^{2}[/tex].
Learn more about the frictional force here:
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Which of the following is the recommended amount of fats per meal for male clients
Answer:
44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.
Explanation:
Answer:
2 thumbs (ISSA Guide)
Explanation:
The students look through the side of the aquarium.
They notice that the image of the tongs appears to break as the tongs enter the water.
Which property of light are the students observing in this situation?
Answer:
light refraction
Explanation:
Answer this question: Is math really important to giving science power? (remember the 5 Ws and the H)
1. The property of a material to absorb liquid like water is called
A non-porous
C. buoyancy
B. density
D. porosity
HELP ME PLEASE I MARK BRAINLIESTIF YOU ANSWER THIS♡
Answer:
A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.
Explanation:
A hygroscopic substance is one that readily attracts water from its surroundings, through either absorption or adsorption.
8) a 20kg box is sliding across the ground. If the coefficient of friction is 0.5, how much friction will the box experience?
A 490 N
B 19.6 N
C 98 N
D 2 N
Answer:
STOP CHETING
Answer:
C is the answer to your question
Explanation:
Have a great day!!! :)
initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity
Answer:
The final velocity is 20 m/s.
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:
[tex]v_f=v_o+at[/tex]
The provided data is: vo=10 m/s, [tex]a=5\ m/s^2[/tex], t=2 s. The final velocity is:
[tex]v_f=10~m/s+5\ m/s^2\cdot 2\ s[/tex]
[tex]v_f=20\ m/s[/tex]
The final velocity is 20 m/s.
If Jack weighs more that Jill, and they run up the same hill, who has done more work?
If Jack and Jill weigh the same, and Jill runs up the hill in half the time as Jack, who had more power?
Answer:
jack has done more work pulling more weight and Jill has more power.
Explanation:
PLEASE ans The question's in the pictures, please don't answer what already has answers. Only answer if you can finish both pages completely PLEASE I NEED HELP :(( if ur ans is relevant I will mark brainliest
The diagram shows an electromagnet made with copper wire, a steel nail,
and a 1.5 V battery. Which action could cause this electromagnet to be
stronger?
A. Replace the steel nail with a plastic straw.
B. Replace the battery with a 6 V battery.
C. Reduce the number of coils of wire wrapped around the nail.
D. Reverse the direction of the battery in the circuit.
Correct answer is B!
Answer:B
Explanation: i took the test and i got it right
Which of the following is true?
A
The Atlantic, Pacific, Indian, Arctic, and Southern Oceans are completely separate
from each other.
B
The ocean covers about half of the Earth's surface.
с
Scientists have studied most of the ocean, but a tiny bit remains unexplored.
D
Scientists know more about the moon than they do the ocean.
Answer:
options B,C,D are true
Explanation:
Any living thing is called an organism,no matter if it is one-celled or many-celled. True or False?.
Answer:
I think it's most likely true.
Explanation:
any organism has the properties of a living thing, which includes cells, whether it has one cell or many
Answer:
False
Explanation:
An organism is a living thing that is a single-celled life form
A horse has a momentum of 1200 kg·m/s. If the horse has a mass of 313 kg, what is the speed of the horse?
Answer:
3.83 m/sExplanation:
The speed of the horse can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]
We have the final answer as
3.83 m/sHope this helps you
If 710-nm and 655-nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.1 m away?
Answer:
Explanation:
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.1 m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2403.07 x 10⁻⁶ m
= 2.403 x 10⁻³ m
= 2.403 mm .
For λ = 655 nm
position = 2 λ D / d
λ = 655 nm , D = 1.1 m
d = .65 x 10⁻³
position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2216.91 x 10⁻⁶ m
= 2.217 x 10⁻³ m
= 2.217 mm .
Difference between their position
= 2.403 - 2.217 = .186 mm .
A solid cylinder is released from the top of an inclined plane of height 0.50 m. From what height, in meters, on the incline should a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?
Answer:
Explanation:
for rolling motion down the plane acceleration is given by the following expression
a = g sinθ / (1 + k² / R²)
here k is radius of gyration and R is radius of the object rolling down .
for cylinder I = 1/2 m R²
so k² = R² / 2
k² / R² = 1/2
a = g sinθ /( 1 + 1 / 2 )
= 2 / 3 x g sinθ
v = √ 2 a s
= √ (2 x 2 / 3 x g sinθ s )
= √ (4 / 3 x g h )
= √ (4 / 3 x g x .5 )
= √ 2g / 3
for sphere I = 2/5 m R²
so k² = 2/5 R²
k² / R² = 2 / 5
a = g sinθ / (1 + 2 / 5)
= 5 / 7 x g sinθ
v = √ 2 a s
= √ (2 x 5 / 7 x g sinθ s )
= √ (10/7 x g h )
Given
√ (10/7 x g h ) = √ 2g / 3
10/7 x g h = 2g / 3
h = 14 / 30 m
= .47 m .
Steelhead trout migrate upriver to spawn. Occasionally they need to leap up small waterfalls to continue their journey. Fortunately, steelhead are remarkable jumpers, capable of leaving the water at a speed of 8.0 m/s. What is the maximum height that a steelhead can jump
Answer:
s = 3.26 m
Explanation:
Given that,
Water leaves at a speed of 8 m/s
We need to find the maximum height that steelhead can jump. Let it can jump to a height of h.
At maximum height, final speed is equal to 0. We can use third equation of motion to find the maximum height.
[tex]v^2-u^2=2as[/tex]
a = -g
[tex]-u^2=-2gs\\\\s=\dfrac{u^2}{2g}\\\\s=\dfrac{(8)^2}{2\times 9.8}\\\\=3.26\ m[/tex]
Hence, the maximum height is 3.26 m.
Thorium^+2
Chemical symbol:
Atomic Number:
Mass: 232
# of protons
# of neutrons
Group #
Period #
Answer:
chemical symbol: Th
atomic number:90
protrons :90
neutrons:142
group#:4
period#: 9
Explanation:
you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons
what instrument is used to measure volume by displacement
PLZ HELP ILL MARK BRAINLEIST!!!!
Why did the bowling ball make a bigger
splash than the ping pong ball?
What kind of energy made that splash
happen?
Answer:
a. The bowling ball would have more kinetic energy because of its greater mass.
b. Potential energy
Explanation
a. Bowling ball has higher mass, self explanatory.
b. A high diver has lots of stored energy when they are on the diving platform. When they dive this stored energy helps make the splash when they hit the water. Stored energy is also called potential energy.
if a person has a mass of 60 kg and a velocity of 2 m/s what is the magnitude of his momentum
Answer:
120 kg m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 60 × 2
We have the final answer as
120 kg m/sHope this helps you
What is the formula for electrical power?
A. P = IV/R
B. P = IVR
C. P = IR
D. P = IV
Correct answer is D!
Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.
What is Power?As we study Electric power, the electric current to the another form of energy.The SI unit of power is the watt or one joule per second.Electric power can be calculated as current times voltage. Electrical energy used by equals the power of the appliances and its multiplied by the amount of time and the appliance is used.Thus, the formula for electric power,
P = IV watt
Where,
P - Power (Watt)
I - Current (Ampere)
V - Voltage (volts)
Power can be also known as the mechanical power equation.Also, the power dissipated energy can also be found with the equation.Electrical power can be defined as the product of Voltage and current. Thus, P = IV. Hence, Option D is the correct answer.
Learn more about Power,
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a 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Find the total momentum of the system.
Explanation:It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :
p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
Answer:
1717 kg*m/s
Explanation:
found it put it in on acellus and it was right
A block with mass m = 0.200 kg is placed against a compressed spring at the bottom of a ramp that is at an angle of 53.0∘ above the horizontal. The spring has 8.00 J of elastic potential energy stored in it. The spring is released, and the block moves up the incline. After the block has traveled a distance of 3.00 m, its speed is 4.00 m/s. Part A What is the magnitude of the friction force that the ramp exerts on the block while the block is moving? Express your answer with the appropriate units.
Answer:
[tex]0.566\; \rm N[/tex] (assuming that while the block is moving, the friction on the block is constant.)
Explanation:
The mechanical energy of a system is the sum of its:
elastic potential energy, gravitational potential energy, andkinetic energy.Friction does work on the block as the block moves up the ramp. The amount of energy that the block-string system has lost would be equal in size to the work that friction has done on the block. The size of the friction on the block could thus be computed.
Before this block was released, the block-spring system has no kinetic energy because there was no movement. Assume that the system has no gravitational potential energy at that moment, either. The only type of mechanical energy in this system at that moment would be elastic potential energy: [tex]8\; \rm J[/tex] according to the question.
The question states that the ramp is [tex]53.0^\circ[/tex] above horizontal. Therefore, after the block has traveled [tex]3.00\; \rm m[/tex] (along the ramp,) the height of this block would have increased by [tex]\Delta h = 3.00\; \rm \sin\left(53^\circ\right) \approx 2.39591\; \rm m[/tex]. Calculate the corresponding gain in gravitational potential energy:
[tex]\begin{aligned}& m \cdot g \cdot \Delta h\\ &\approx 0.200\; \rm kg& \\\ &\quad\times 9.81\; \rm N \cdot kg^{-1} \\ &\quad \times 2.39591\; \rm m \\ &\approx 4.70077\; \rm J\end{aligned}[/tex].
On the other hand, the question states that the speed of the block ([tex]m = 0.200\; \rm kg[/tex]) at that moment is [tex]v = 4.00\; \rm m \cdot s^{-1}[/tex]. Calculate the corresponding kinetic energy:
[tex]\begin{aligned}& \frac{1}{2}\, m \cdot v^{2} = 1.6\; \rm J \end{aligned}[/tex].
Because the block was already released, there should be no elastic potential energy in the spring.
Hence, the mechanical of the block-spring system would be approximately [tex]4.70077\; \rm J+ 1.6\; \rm J \approx 6.30077\; \rm J[/tex].
Approximate the amount of mechanical energy that is lost:
[tex]8\; \rm J - 6.30077\; \rm J \approx 1.69923\; \rm J[/tex].
In other words, when applied over [tex]3\; \rm m[/tex], the friction on this block would do approximately [tex]1.69923\; \rm J[/tex] of work. Approximate the size of that friction:
[tex]\begin{aligned}F &= \frac{W}{s} \\ &\approx \frac{1.69923\; \rm J}{3.00\; \rm m}\approx 0.566\; \rm N\end{aligned}[/tex].
A rocket falls from the apogee (0 meters per second) until it hits the ground with a speed of 10 meters per second. Gravity pulled it down with an acceleration of 9.8m/s^2. The time during which the ball is in free fall is approximately what time?
Answer:
Approximately 1.02 seconds
Explanation:
Use the final velocity (vf) formula for a uniformly accelerated movement under "g" (acceleration of gravity):
[tex]v_f=v_i+g\,*\,t[/tex]
in our case:
[tex]10=0+9.8\,*\,t\\t=10/9.8\\t\approx 1.02\,\,sec[/tex]
While getting buff at the gym you lift a bunch of weights applying 1000N of force to lift them from the ground to a height of 2m. How much work did you do?
A. 2000 J
B. 1000J
C. -2000 J
D. -1000 J
The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .
Answer:
It’s a
Explanation:
Don’t actually put that i needed the points mb
A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
Answer:
20 m/s^2
Explanation:
Use the formula Vf=Vi+at
Vf=100
Vi=0
a=?
t=5
Then solve for (a)
A 80 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 5 meters starting from rest, its speed is 6.0 m/s. Find the magnitude of the net force on the bobsled.
How do you solve this question?
Answer:
F = 288 [N]
Explanation:
To solve this problem we must use the following equation of kinematics and find the value of acceleration.
[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 6 [m/s]
Vo = initial velocity = 0 (starting from rest)
a = acceleration [m/s²]
x = distance = 5 [m]
Now replacing, we have:
[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]
Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
ΣF = m*a
[tex]F =80*3.6\\F = 288 [N][/tex]
A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.
-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?
Answer:
a)t = 1,43 s
b) V = 10,49 m/s
c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s
d) Vf = 17,5 m/s
Explanation:
According to the problem statement
V₀ = V₀ₓ and V₀y = 0
And at the end of the movement t = ? the distance y = 10 m
Therefore as
h = V₀y - (1/2)*g*t²
Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²
10 = 4,9*t²
t² = 10/4,9 ⇒ t² = 2,04 s
t = 1,43 s
a) 1,43 s is the time of movement
b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )
Just before touching the ground, the horizontal distance is
hd = 15 = Vₓ * t
Then 15 /1,43 = Vₓ = V₀ₓ
Vₓ = 10,49 m/s
Then initial speed is V = 10,49 m/s since V₀y = 0
Vf² = Vₓ² + Vy²
Vyf = V₀y - g*t
Vyf = 0 - 9,8 *1,43
Vyf = - 14,01 m/s
And finally the speed when the projectile strike the ground is:
Vf² = Vₓ² + Vy²
Vf = √ (10,49)² + (14,01)²
Vf = 17,50 m/s