Answer:
Explanation:
Let when cat reaches the point L distance from the right end , the plank starts turning about the left sawhorse.
The weight of the plank will at at middle point , ie at 2 m from the right end .
The plank will turn about the left sawhorse . Taking torque about this point of weight of the plank and weight of the cat , we have
2.5 g x ( L - 2.4 ) = 6 x g x ( 2.4 - 2.0 )
2.5 L - 6 = 14.4 - 12
2.5 L = 8.4
L = 3.36
position from left sawhorse = 3.36 - 2.4 = .96 m to the left of left sawhorse.
A 6.0 kg object is moving at 5.0 m/s along the x axis in the positive direction. It collides
with and sticks to a 2.0 kg object moving also along the x axis. After the collision the
composite object is moving 2.0 m/s along the x axis in the negative direction.
Determine the velocity (magnitude and direction) of the 2.0 kg object before the collision.
Answer:
V2 = 23 [m/s] to the left.
Explanation:
In order to solve this problem, we must use the definition of conservation of linear momentum. That is, the momentum is conserved before and after the collision. The values before the collision will be taken to the left of the equality, and the values after the collision will be taken to the right of the equality, in this way we have:
Σbefore = Σafter
ΣPbefore = ΣPafter
where:
P = m*v
The positive momentum will be taken to the right and the negative momentum is to the left in this way we formulate the following equation:
[tex](m_{1}*v_{1}) + (m_{2}*v_{2})=-(m_{1} +m_{2})*v_{3}\\[/tex]
where:
m1 = mass of the first object = 6 [kg]
v1 = velocity of the first object = 5 [m/s]
m2 = mass of the stick = 2 [kg]
v2 = velocity of the stick [m/s]
v3 = velocity of the composite object = - 2 [m/s]
(6*5) + (2*V2) = - (6 + 2)*2
30 + (8*2) = - 2*V2
46 = - 2*V2
V2 = - 23 [m/s]
Note: the negative sign means the stick moves to the left
onsider what happens when you jump up in the air. Which of the following is the most accurate statement?A) Since the ground is stationary, it cannot exert the upward force necessary to propel you into the air.Instead, the internal forces of your muscles acting on your body itself propels the body into the air.B) The upward force exerted by the ground pushes you up, but this force can never exceed your weight.C) When you jump up the earth exerts a force F1on you and you exert a force F2 on the earth. You go upbecause F1 > F2, and this is so because F1 is to F2 as the earth's mass is to your mass.D) You are able to spring up because the earth exerts a force upward on you which is stronger than thedownward force you exert on the earth.E) When you push down on the earth with a force greater than your weight, the earth will push back with thesame magnitude force and thus propel you into the air.
Answer: D
Explanation: it seem right to me I really don't know if this right but I hope this helps
Help real quick someone
I need to know what the answer is to this
Answer:
i think its the top one
Explanation:
pls tell me if im wrong
If object A exerts a force on another body B, then body B exerts an equal and opposite for on body A. Which newton is this?
Answer:
newton 's 3rd law which states that to every action there's equal but opposite reaction
If you were to pour 4 different liquids into a glass and the liquids separated into 4 layers, which liquid would have the highest density?
A. the layer at the bottom of the glass
B. the third layer from the bottom of the glass
C. the layer at the top of the glass
D. the second layer from the bottom of the glass
Answer:
A.
Explanation:
The liquid with the higher density will be found at the bottom of the glass because is the heaviest of all. The liquids which have a lower density than the liquid at the bottom of the glass will be found at the top of the glass. We assume that the liquids are not soluble in each other because otherwise they will be mixed.
e.) calculate the total heat required to change 10kg of ice at 0°c into water at 50°c.
Answer:
the heat change to 10kg of ice to water 0
so quantity of heat required is answer: 5460 J.
hope its helps!
Answer:
Q=ml+mc∆+ml'
=10*80+10*1*(100-0)+10*540
=800+1000+5400
=7200cal.7.2kcel
L=heat of fusion of ice
L'=heat of vapourisation of water
what is the order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20 meters per second2 for 5.0 seconds
The order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s
The acceleration of a body is the change in velocity with respect to time as shown:
[tex]a=\frac{v-u}{t}[/tex]
Given the following parameters
a is the acceleration = 20m/s²
u is the initial velocity = 0m/s
t is the time taken = 5.0seconds
Required
Final velocity "v"
Substitute the given parameters into the formula:
[tex]20=\frac{v-0}{5} \\20 =\frac{v}{5}\\v = 20 \times 5\\v =100m/s[/tex]
Hence the order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s
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describe an example that shows thermal energy cannot be stored for a long time
Answer: heat always flows from higher temperature to lower temperature
Explanation: there:)
Heat always flows from higher temperature to lower temperature so can not be stored.
What is thermal energy?The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is performed in the process.
Heat dissipation always happen decrease in temperature so thermal energy can not be stored for long time.
Heat always flows from higher temperature to lower temperature so can not be stored
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A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
submerged in a liquid with SG = 1.59.
Answer:
8.8 kN
Explanation:
V = 2 m³, W = 40 kN, SG = 1.59
Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN
So the weight becomes 40 - 31.2 = 8.8 kN
Which information did the Glomar Challenger study in 1968?
the rate of seafloor spreading
the direction of seafloor spreading
the age of rocks in various places in the ocean
the contents of rocks in various places in the ocean
Explanation:
Glomar Challenger studies about the "age of rocks in various places in the ocean" in 1968. EXPLANATION: Glomar Challenger was a "deep sea research vessel" for marine geology and oceanography studies.
I hope this helps you :)
Answer:
c no cappp :)
Explanation:
True or false: humans have to find a balance with their environment, using sparingly so we don’t run out of them
Answer:
False
Explanation:
in my point of view the human race has adapted to being greedy in some ways we can save resources but its very rare
hope this helps! : )
Energy from food
(Choose all that are correct )
Can be stored for later
Can be used to keep your heart beating
Can keep you warm
All of the above
Answer:
all of the above I think??
At low pressures and high temperatures, the density of a gas
Answer:
Higher denisty
Explanation:
High pressure=high denisty
A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed at the bottom of a plane?
Answer:
V2 = 15.53 [m/s]]
Explanation:
In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.
Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.
[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]
A force of 3N acts on 90degree to a force of 4N.find the magnitude and direction of the resultant R.
R=A2+B2+2ABcosβ−−−−−−−−−−−−−−−−√R=A2+B2+2ABcosβ
A=4NA=4N , B=3NB=3N , β=90°β=90° , cosβ=0cosβ=0
R=A2+B2−−−−−−−√R=A2+B2
R=42+32−−−−−−√=25−−√=5NR=42+32=25=5N
tanα=Bsin90°A+Bcos90°=34tanα=Bsin90°A+Bcos90°=34
α=37°α=37°
Therefore the resultant of the two forces has a magnitude of 5N5N and is at an angle of 37°37° with respect to
The magnitude of the vertical velocity vector for an upwardly launched projectile _________. a stays constant b gets smaller and then larger c decreases throughout the flight d increases throughout the flight
Answer:
changes by 9.8 m/s each second.
is water wet? If water is not wet does that make it dry?
Answer:
Water isn't wet by itself, but it makes other materials wet when it sticks to the surface of them.
Explanation:
Answer:
water is wet
it is a liquid
Explanation:
An image of the Earth-moon-sun system is shown.The moon remains in orbit around Earth because of the force of —
A student claims that the cart can only be at rest if it is experiencing balanced forces. Choose the statement that correctly evaluates this student’s claim.
A.) The student is correct because only objects that are experiencing unbalanced forces can move.
B.) The student is correct because an object experiences balanced forces will remain at rest. C.) The student is incorrect because experiencing unbalanced forces can cause an object to come to rest.
D.)The student is incorrect because objects moving at a constant velocity is experiencing balanced forces.
Kepler's work revealed that the Earth was at the center of the circular orbits of the planets at the center of the elliptical orbits of the planets orbiting the Sun in an elliptical orbit O at one focus of the elliptical orbit of the planets
Answer:
orbiting the Sun in an elliptical orbit
Explanation:
When orbits are circular, the center and the focus are the same point. For an elliptical orbit, the orbited object is at one focus of the ellipse. Kepler found planetary orbits to be elliptical with the sun at one focus.
__
As with a lot of scientific theories, it only explained some of the motion. As with a lot of scientific theories, explaining the discrepancies resulted in new discoveries.
Answer:
at one focus of the elliptical orbit of the planets.
1. Allen is driving North on Highway 69 at 90 km/h and sees a large moose on the road. He
quickly slams on his brakes, but his reaction time is 0.85 s (as he sees the moose, thinks
about his response, and then presses the brake pedal). He presses the brake for 3.5 s and
comes to a stop just in time.
a) Find the distance travelled after seeing the moose and before pressing the brake.
b) Find the total distance he travelled before coming to a stop.
c) Find the average acceleration once he presses the brake.
Take the moment Allen sees the moose to be the origin.
First, convert his speed to m/s.
90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s
(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of
(25 m/s) • (0.85 s) = 21.25 m
(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of
x = 21.25 m + (25 m/s) t + 1/2 a t²
and has a speed v of
v = 25 m/s + a t
It takes him 3.5 s to come to a full stop. Use this to find the acceleration:
0 = 25 m/s + a (3.5 s)
a = - (25 m/s) / (3.5 s)
a ≈ - 7.1 m/s²
After 3.5 s, he will have traveled a total distance of
x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²
x = 152.5 m ≈ 150 m
(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be
a ≈ - 7.1 m/s²
Calculate the wave speed (in m/s) for the following waves:
a) A sound wave in steel with a frequency of 500 Hz and a wavelength of 3.0 meters. (2pts)
b) a ripple on a pond with a frequency of 2 Hz and a wavelength of 0.4 meters. (2pts)
Calculate the wavelength (in meters) for the following waves:
A wave on a slinky spring with a frequency of 2 Hz travelling at 3 m/s. (2pts)
An ultrasound wave with a frequency 40,000 Hz travelling at 1450 m/s in fatty tissue. (2pts)
Calculate the frequency (in Hz) for the following waves:
A wave on the sea with a speed of 8 m/s and a wavelength of 20 meters. (2pts)
A microwave of wavelength 0.15 meters travelling through space at 300,000,000 m/s. (2pts)
Answer: A : 250 is the answer
B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.
Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.
In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is
where is called Hertz (Hz). So, the correct answer is
Explanation:
#Wavespeed
#1
[tex]\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s[/tex]
#2
[tex]\\ \rm\Rrightarrow v=2(0.4)=0.8m/s[/tex]
#Wavelength
#1
[tex]\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m[/tex]
#2
[tex]\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m[/tex]
#Frequency
[tex]\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz[/tex]
#2
[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz[/tex]
how does gravity change as it nears an object
SAVE MEEEE WILL MARK BRAINLY
How much time would it take for a 0.17 kg ice hockey puck to decrease its speed by 9.0 m/s if the coefficient of kinetic friction between the ice and the puck is 0.05
How much time would it take for a 0.17 kg ice hockey puck to decrease its speed by 9.0 m/s if the coefficient of kinetic friction between the ice and the puck is 0.05
Given:- mass = 0.17 kg change in speed= 9 m/scoefficient of kinetic friction = 0.05 To Find :-Time taken to decrease the speed
Answer:-Equation :-
[tex]f {\tiny{k} }= u {\tiny{k}}.f \tiny{N}[/tex]
[tex]f {\tiny{N} }= mg = 0.17 \times 10 \\f {\tiny{N} }= 1.7 {N}[/tex]
[tex]u{ \tiny{N}} = 0.05[/tex]
[tex]mg = f {\tiny{k} }= u {\tiny{k}}.f {\tiny{N}} \\ 0.17a = 0.05 \times 1.7 \\ a = 0.05 \times \frac{ \cancel{1.7} {}^{ \: \: 10} }{ \cancel{0.17}} \\ a = 0.5m {s}^{ - 2} [/tex]
change in speed = 9 ms-1 (Given)
[tex]change \: in \: speed = at \\ 9 = 0.5 \times t \\ \frac{9 \times 10}{5} = t \\ \frac{9 \times \cancel{10} {}^{ \: \: 2} }{ \cancel{5}} = t \\ 18 \: sec = t \: [/tex]
What is an independent variable?
A ball is travelling 32° above the horizontal at a speed of 24 m/s. What is the horizontal component of its speed
A. 12.7 m/s
B. 13.0 m/s
C. 29.2 m/s
D. 20.4 m/s
Answer:
Since the ball is travelling 32 degrees above the horizontal, the value of Θ is 32
In the figure, v vector is the vertical component whereas h vector is the horizontal component
Using trigonometry:
CosΘ = h /24
Cos 32 = h/ 24
0.85 = h / 24
h = 24*0.85
h = 20.4 m/s
Describe the buoyant force and explain how
it relates to Archimedes principle.
Answer:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
Please I need help with this :(
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
2. Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m against a force of 23 N.
Answer: 6 J
Explanation:
Total force applied = 47 N Assuming that direction of movement of pencil and applied force is same. Work done by force in moving the pencil W = Force × Distance through which force moves ⇒ W = 47 × 0.25 = 11.75 J .-.-.-.-.-.-.-.-. In case we are asked useful work done then we calculate net force used for pushing the pencil: Net force used for pushing the pencil = 47 − 23 = 24 N Assuming that direction of movement of pencil and net force is same. Useful Work done by force in moving the pencil W u = Force × Distance ⇒ W u = 24 × 0.25 = 6 J
The work done the pushing force is required.
The work done by the pushing force is [tex]6\ \text{J}[/tex]
[tex]F_1[/tex] = Pushing force = 47 N
[tex]F_2[/tex] = Opposing force = 23 N
m = Mass of object = 0.025 kg
s = Displacement = 0.25 m
Work done is given by
[tex]W=F_ns[/tex]
[tex]\Rightarrow W=(47-23)\times 0.25[/tex]
[tex]\Rightarrow W=6\ \text{J}[/tex]
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