Answer:it could be B
Explanation:
im not sure
I need it in the next hour or so!
A sports car with a mass of 650 kg is accelerating at +3 m/s2. The force of friction is 200 N, the forward force from the engine is 3100 N. What must the force of air resistance on the car be? - Show your work - Include a unit with your answer
The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
Can someone please help me with number 10
Answer:
A
Explanation:
increase
Answer:
B???
Explanation:
The rate of it going up is lower than L
It might be D too.
Can someone please give me the (Answers) to this? ... please ...
Answer:
Jake’s horse will only need to exert enough force to make the carriage accelerate. Even though the carriage will pull on the horse as well, that force is not enough to make the horse accelerate in the direction of the carriage because it is only strong enough to make the carriage accelerate, not the horse.
Explanation:
Additional info: the horse has more mass than the carriage does so it would require a stronger force to make the horse accelerate.
Answer: Jake’s horse will only need to exert enough force to make the carriage accelerate .Since the horse has a greater mass than the carriage, it requires a stronger force to make the horse accelerate.
Explanation: Sorry it took so long lol
Explain Body Mechanics.
PLUS BRAINLY! and 50 Points
Answer Body mechanics is a term used to describe the ways we move as we go about our daily lives. It includes how we hold our bodies when we sit, stand, lift, carry, bend, and sleep. ... Good body mechanics means using the body's strength to the best mechanical advantage to do a task efficiently and without injury.
Explanation:
What are the three different social perspectives on sport
Answer:
functionalist theory,feminist theory. discipline of sociology
A 78 N block is supported by a spring whose constant is 12 N/m. Calculate the elongation of the spring under this load.
a. 6.5 m
b. 1.5 m
c. 8.2 m
d. 7.2 m
Answer:
a 6.5 m
Explanation:
[tex]f = kx = > x = \frac{f}{k} = \frac{78}{12} = 6.5 m[/tex]
What type of heat does not require matter?
Radiation is the only method of heat transfer that does not require matter.
Facts about Radiation :
The energy from the sun is received in the form of radiation. It travels in the form of waves. It can be absorbed by substances in its path.Heat Radiation travels through air. Black surfaces are the emitters of heat radiation.Shiny surfaces are the reflectors of heat radiation.Heat transfer by Radiation doesn't require any medium.Which statement about diffraction is correct?
O The amplitudes of two waves combine to appear as one big wave.
O Sound waves bend around the corners of various obstacles.
O The amplitudes of two waves combine to appear as a wave smaller than the individual waves.
O Sound waves can only travel in straight lines.
Answer:
Sound waves bend around the corners of various obstacles.
Explanation:
I took the test
Liquids and solids are:
a- compressible and expandable.
b- incompressible and non-expandable.
c- incompressible and expandable.
Answer:
incompressible and expandable
A rock is lying on a rock ledge that is 3 m high. The rock as 120 J of potential energy. What is the mass of the rock?
Have a great day!
[tex]\\ \sf\Rrightarrow PE=mgh[/tex]
[tex]\\ \sf\Rrightarrow m(10)(3)=120[/tex]
[tex]\\ \sf\Rrightarrow 30m=120[/tex]
[tex]\\ \sf\Rrightarrow m=4kg[/tex]
Mass=4kgBlock 1, of mass m1
m
1
= 0.500 kg
k
g
, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2
m
2
, as shown. For an angle of θ
θ
= 30.0 ∘
∘
and a coefficient of kinetic friction between block 2 and the plane of μ
μ
= 0.400, an acceleration of magnitude a
a
= 0.450 m/s2
m
/
s
2
is observed for block 2.
Find the mass of block 2, m2
m
2
.
Answer:
M2=0.52kg
Explanation:
thats rightt
A horizontal force is applied to a box with a mass equal to 5.5 kg, The graph shows the net force acting on a box as a function of its horizontal position x.
Part A
Find the net work done on the box as it moves from x = 0 m to x = 8.0 m. .
Part B
Apply concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m. Assume the box started from rest.
From the graph, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
From the question, these are the given parameters;
Mass M = 5.5 KgPosition X = 0 to X = 8mPart A
The net work done will be the area under the graph.
From position X = 2m to X = 8m gives us the shape of a trapezium.
A = 1/2( a + b )h
A = 1/2( 2 + 6 ) x 8
A = 8 x 4
A = 32 Nm
From X = 0 to X = 2 gives us the shape of a triangle.
A = 1/2bh
A = 1/2 x 2 x (-4)
A = -4 x 1
A = -4 Nm
Net Work done = 32 - 4
Net work done = 28 Nm
Part B
Applying the concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m
Net Work done = 1/2m[tex]V^{2}[/tex]
Substitute all the necessary parameters
28 = 1/2 x 5.5 x [tex]V^{2}[/tex]
5.5[tex]V^{2}[/tex] = 56
[tex]V^{2}[/tex] = 56/5.5
[tex]V^{2}[/tex] = 10.18
V = [tex]\sqrt{10.1818}[/tex]
V = 3.19 m/s
Therefore, net work done on the box and the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively
Learn more about work done here: https://brainly.com/question/8119756
Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.
Position (km) Time (h)
0.0 0.000
1.2 0.024
2.8 0.051
4.2 0.084
16.3 2.340
The average speed between 0 h and 2.340 h is 6.97 Km/h
Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.
[tex]Average \: speed = \frac{total \: distance}{total \: time} \\ \\ [/tex]
With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:
Total time = 2.340 – 0 = 2.340 hTotal distance = 16.3 – 0 = 16.3 KmAverage speed =?[tex]Average \: speed = \frac{total \: distance}{total \: time} \\ \\Average \: speed = \frac{16.3}{2.340} \\ \\ Average \: speed = 6.97 \: Km/h \\ \\ [/tex]
Learn more about average speed: https://brainly.com/question/24884027
Answer:
6.983 km/hr
Explanation:
The average rate of change is (16.34 - 0.0)/(2.34 - 0.000) ≈ 6.983 km/hr
Mejor yo por favor
Help me as soon as possible <3
Answer: ??? what the
Explanation:
Resultant of two froce 3N nad 4N acting ar apoit is 5N show that the two froce are 2 perpendicular to each other
[tex]\qquad[/tex] ☀️We are asked to prove that two forces are perpendicular to each other.
Formula to calculate magnitude of resultant 'R' of two forces 'A' and 'B' when they are at an angle 'p' to each other.
[tex]\qquad[/tex][tex]\purple{ \longrightarrow \bf R^2= A^2 + B^2 + 2 AB cos p} [/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf 5^2= 3^2 + 4^2+ 2 \times 3\times 4\times cos p[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf 25 = 9 + 16 + 24 cos p [/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf 25 = 25 + 24 cos p[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf 25 - 25 = 24 cos p[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf 0 = 24 cos p[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf cos p = 0[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf cos p = cos 90°[/tex]
[tex]\qquad[/tex][tex]\purple{ \longrightarrow \sf p = 90°}[/tex]
Hence, Proved that –
Forces A = 3N and B = 4N are Perpendicular to each other.[tex]\qquad[/tex]_________________________
A student holds an egg outside of an open window and let's go. The window is 40 meters above the ground and the egg is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
How far will the egg travel after 2 seconds? Be sure to use the 5 steps to solve the problem. Show which equation you used, what each variable equals, and answer the question in a complete sentence.
The distance traveled by the egg after 2 seconds is 19.6 m.
The given parameters:
Height of the window, h = 40 mThe distance traveled by the egg after 2 seconds is calculated as follows;
[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\[/tex]
where;
[tex]v_0_y[/tex] is the initial vertical velocity of the egg = 0g is acceleration due to gravity = 9.8 m/s²t is the time of motion, = 2 s[tex]h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m[/tex]
Thus, the distance traveled by the egg after 2 seconds is 19.6 m.
Learn more about second equation of motion here: https://brainly.com/question/7977693
An instrument made of brass is an example of which of these?
solid–liquid solution
solid solution
liquid–liquid solution
gas–liquid solution
Answer:
a solid made of liquid solution
Explanation:
a trumpet for example has intrcuite valves , the only way to do this is by creating a liquid form of the metal to be able to create the solid formed instrument.
Answer:
a solid made of liquid solution
Explanation:
domne
A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what are the X AND Y components of the NORMAL FORCE?
Answer:
y-component = 45.75N
x-component = 0N
Explanation:
Was there any angle given with the question
Answer:
No.
Explanation:
In the question, no mathematical angle is shown.
What happens to a liquid in a U-tube manometer when placed in a vacuum
Answer:
With a greater pressure applied to the left side of a U-tube manometer, the liquid lowers in the left leg and rises in the right leg. The liquid moves until the unit weight of the liquid, as indicated by h, exactly balances the pressure.
Answer:
Here
Explanation:
With a greater pressure applied to the left side of a U-tube manometer, the liquid lowers in the left leg and rises in the right leg. The liquid moves until the unit weight of the liquid
How to do this question plz
Answer:
Let L be the length of the wire.
N λ = M L where N is the number of wavelengths λ. Also, M and are integers and L is the length of the wire
There must be nodes at the ends of the wire and anti-nodes at the points of zero displacement - λ/2, λ, 3λ/ 2, etc.
The simplest would be 2 λ = L or L = λ / 2
In such a situation one would have
N-A-N or node-antinode-node
There is a 180 degree phase change upon reflection of the wave creating the standing wave
The distance between nodes or anti-nodes must be λ/2 and there must be nodes at the ends of the wire
HELP!!!!
Which statement describes earthquakes?
They release energy.
They are caused by reduced stress in rocks.
They begin at the epicenter.
They result from movement on Earth’s surface.
Answer:
A
Explanation:
earthquakes release energy
Answer:
i think it begins at the epicenter
Explanation:
i can remember
pls give brainleist
a 280 nm thin film with index of refraction 1.6 floats on waterwhat is the largest wavelength of reflected light for which constructive interference occur? answer in units of m.
Answer:
Inside the film the wavelength will be λ/n
For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur
280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed
Note that 700 nm is also the upper wavelength of the visible spectrum
The Sun has a mass of 1.99 x 10^30kg [^30 is an exponent] and is 1.5 x 10^11m [^11 is an exponent] from the earth. The planet Earth is 5.98 x 10^24kg [^24 is an exponent]. What is the gravitational attraction between the sun and the earth? G=6.67×10^-11 (-11 is an exponent)
We are given –
Mass of earth,m = 5.98×10²⁴ kgMass of sun, M= 1.99×10³⁰ kgDistance between Earth andSun, r= 1.5×10¹¹ m
G = 6.67×10⁻¹¹ Nm²/kg²We are asked to find the gravitational attraction between the sun and the earth.
By Universal Law of Gravitation, we know that –[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\sf F = \dfrac{GMm}{r²}}[/tex]
Substituting values, we get –
[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf F = \dfrac{6.67×10^{-11}\times 5.98×10^{24}\times 1.99×10^{30}}{(1.5×10^{11})^2}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf F = \dfrac{6.67×10^{-11}×5.98×10^{24}×1.99×10^{30}}{2.25×10^{22}}[/tex]
[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf F = 35.27×10^{21}[/tex]
[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf F = 3.53×10^{22} N}[/tex]
Henceforth ,gravitational attraction between the sun and the earth. is 3.53×10²²N .______________________________________________
A 263 g ball of clay falls onto a vertical spring.
The spring constant k = 2.52 N/cm.
The clay sticks to the spring and as the spring compresses a distance of 11.8 cm the clay momentarily comes to rest.
Determine the speed of the clay just before it hit the spring in [m/s]?
As the spring is compressed, it performs
-1/2 (252 N/m) (0.118 m)² ≈ -1.75 J
of work on the clay, while gravity does
(0.263 kg) (9.80 m/s²) (0.118 m) ≈ 0.304 J
on it. Then the total work W performed on the clay ball is approximately -1.45 J.
By the work-energy theorem, W is equal to the change in the clay ball's kinetic energy ∆K. If v is the speed with which it hits the spring, then
W = ∆K
-1.45 J = 0 - 1/2 mv²
Solve for v :
-1.45 J = -1/2 (0.263 kg) v²
v² ≈ 11.0 m²/s²
v ≈ 3.32 m/s
Look at the diagram of two trains. Calculate the relative speed.
Answer:speed difference
Explanation:
compared to spiral galaxies, elliptical galaxies are:
flat disk like shape
________________
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how to find change in kinetic energy wfrom joules
Answer:
What do you mean? Can you please explain yourself better?
the normal formula to find force is F = m*a. What kind of math do you need to do
to find the mass of an object?
- Divide: F / a
- Divide: a / F
- Multiply: F* a
- Subtract: F - a
[tex]\\ \sf\Rrightarrow F=ma[/tex]
Take a to leftAs it's multiplied on right side it will be divided on right side.[tex]\\ \sf\Rrightarrow \dfrac{F}{a}=m[/tex]
Or
[tex]\\ \sf\Rrightarrow m=\dfrac{F}{a}[/tex]
Chapter name :- Physical Optics
Question :-
When two waves of magnitude I and 4I overlap, the maximum and minimum light intensity will be-
a) 9I, 3I
b) 3I,I
c) 9I, I
d) 5I, 3I
Guys I need the correct answer in details!
✖️❌Please No links, No rubbish answer! Otherwise, I'll report your answer!
Answer:
Hey There!
Refer to the attachment...
or in this way also you can solve shown in second attachment
I hope it is helpful to you...Cheers!_________