Answer:
i think it's C thx correct me if wrong
If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must be true of the magnitude of the reaction force?
Newton's third law allows to find the result for the value of the reaction force during the collision is:
The reaction force is F = 90 N and is applied to the lighter body.Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.
These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.
They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.
Consequently, the reaction force is F = 90 N directed towards the lighter body.
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If the velocity and frequency of a wave are both doubled, how does the wavelength change?
The wavelength will remain unchanged.
Explanation:
The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is
[tex]v = \lambda\nu[/tex] (1)
so if we double both the velocity and the frequency, the equation above becomes
[tex]2v = \lambda(2\nu)[/tex] (2)
Solving for the wavelength from Eqn(2), we get
[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]
We would have gotten the same result had we used Eqn(1) instead.
Answer:
the wavelength increases
Explanation:
The mass of a car is 625kg. Calculate the weight of the gravitational field strength is 10 N/kg.
Ans: 62.5
Explanation: [tex]F{net}[/tex] = m x a
1N = 1kg x 1m/ [tex]s{2}[/tex]
Understanding what motivates anyone is not easy because each individual has different
48.36
g.
MgSO4 to motes
Answer:120.3676
Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!
What is the approximate value of k when 30 = e^5k?
Answer:
Explanation:
30 = e^5k
ln30 = lne^5k
ln30 = 5k
k = ln30/5
k = 0.68023947...
round to your heart's content.
A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.
Answer:
a = 0.01m/s²
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
a = (V_f-V_0)/t
a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))
a = 0.01m/s²
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
Answer:
Explanation:
The work increased the potential energy
W = PE = mgh = 40(9.8)(15) = 5880 J(oules)
is it true that playing badmenton help you to become a better person?
Answer:
There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.
Answer:
It's true because playing any sport makes a person happy. So a happy person is a better person.
Please Mark as brainliest.
in a compoumd are atoms physically or chemically combined
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
The qualitative equivalent of external validity is:
A- Credibility
B- Dependability
C- Transformability
D- Confirmability
I need help. please look at the image below and let me know I need this by 7:20 am pst.
Answer:
3(1.5) = 4.5 V
Explanation:
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
what type of stretching is beneficial for sports performance and involves momentum
Answer:
Dynamic stretching
Explanation:
Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.
A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?
[tex]a_c = 3.14\:\text{m/s}^2[/tex]
Explanation:
First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:
[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]
The centripetal acceleration [tex]a_c[/tex] is defined as
[tex]a_c = \dfrac{v^2}{r}[/tex]
Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as
[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]
[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]
a stone is thrown down off a bridge with a velocity of 22 m/s. what is its velocity after 1.5 seconds has passed?
Answer:
Velocity of the stone after 1.5 seconds has passed = 37 m/s
Explanation:
Initial velocity (u) = 22 m/s
Time (t) = 1.5 sec
Acceleration due to gravity (g) = 10 m/s²
By using kinematics equation:
v = u + gt
v = 22 + 10 × 1.5
v = 22 + 15
v = 37 m/s
Final velocity (v) = 37 m/s
Please help me.............................
Answer:
[tex]a[/tex]
Explanation:
is a corect anser
Help me outtttt jejjejejeje
Answer:
do it got a picture
Explanation:
You are angry at Dr. Anderson for this exam, so you throw a 0.30-kg stone at his car with a speed of 44 m/s. How much kinetic energy does the stone have
Answer:
Explanation:
KE = ½mv²
KE = ½(0.30)44²
KE = 290 J rounded to 2 s.d.
A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m
Answer:
P = 2 pi (L / g)^1/2
P2 / P1 = (8 / 2)^1/2 = 2
The period would be twice as long or 5.6 sec.
A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid
The density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
What is density?The density of an object is defined as the ratio of the mass of an object to the volume of the object.
Volume of tube = 2^2 * pi * 30 = 377 cm^3
Volume of tube submerged = 25* 377 / 30 = 314 cm^3
Buoyancy = weight of liquid displaced
Volume of liquid displaced = 314 cm^3
Mass of tube and lead = 250 + 30 = 280 g
Now from the mass density by definition
[tex]\rho = \dfrac{m}{v}[/tex]
[tex]m=\rho \times v[/tex]
Mass of liquid displaced = Mass being supported
[tex]314 \times \rho = 280 g[/tex]
[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]
Thus the density of the liquid will be equal to [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]
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The density of the liquid is 1.67 g/[tex]cm^3[/tex].
The volume of the tube is
30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].
The mass of the lead pellets and the plastic tube is
30 + 250 = 280 g.
The volume of the lead pellets is
250 / 11.34 = 22 [tex]cm^3[/tex].
The volume of the liquid that the tube displaces is
94.2 - 22 = 72 [tex]cm^3[/tex].
The density of the liquid is
280 / 72 = 1.67 g/ [tex]cm^3[/tex].
Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].
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A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m
Answer:
Explanation:
f = [tex]\sqrt{T/(m/L)} / 2L[/tex]
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = [tex]\sqrt{120/12} /(2(3))[/tex])
f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.
Hi there!
Recall Newton's Law of Universal Gravitation:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Where:
Fg = Force of gravity (N)
G = Gravitational Constant
m1, m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the given values stated in the problem:
[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]
Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly
The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
EQUILIBRIUM:A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.
In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.
Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.
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An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)
0.8 seconds
Explanation:
time of flight = 2u/g
u=4m/s
g=10
= 8/10
= 0.8 sec
just a trial...not sure!!!
Given :
∅ = 60⁰
u = 4 m/s
g = 10m/s²
to find :
T = ?
Solution :
as per formula,
[tex]t = \frac{2u \: sin \theta}{g} [/tex]
now put the value : [tex]t \: = \frac{2 \times 4 \times sin \: 60}{10} [/tex]
as we know [tex] sin60 \: = \frac{ \sqrt{3} }{2} [/tex]
therefore,
[tex]t \: = \frac{8 \times \frac{ \sqrt{3} }{2} }{10} [/tex]
as we solve this we get,
[tex]t \: = \frac{ 2\sqrt{3} }{5} [/tex]
that's t = 0.69 sec
[tex]\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}[/tex]
3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant velocity for 30s and come to rest with uniform acceleration of 2m/s2. Calculate the total distance covered by the car.
The total distance traveled by the car at the given velocity and time is 900 m.
The given parameters:
initial velocity of the car, u = 20 m/sacceleration of the car, a = 12 m/s²time of motion of the car, t = 20 sfinal time = 30 sfinal acceleration = 2 m/s²The final time of motion of car before coming to rest is calculated as follows;
[tex]v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s[/tex]
The graph of the car's motion is in the image uploaded.
The total distance traveled by the car is calculated as follows;
[tex]total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m[/tex]
Thus, the total distance traveled by the car at the given velocity and time is 900 m.
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12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]
Hi there!
We can use the following:
W = ΔKE = F · d
Find the work done on the cart:
W = 200 · 10 = 2000 J
Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:
2000J = KEf - KEi
KE is given as:
[tex]KE = \frac{1}{2}mv^2[/tex]
2000J = 1/2(55)v²
4000 = 55v²
√(4000/55) = 8.53 m/s
A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?
What height will the football reach?
With what speed does the punter need to kick the football?
At what angle (θ), with the horizontal, does the punter need to kick the football?
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.
Answer:
Weight of object = 11.2 N
Apparent weight = 3.83 N when immersed
Weight of water displaced = 11.2 - 3.83 = 7.37 N
d (density) W / V weight / volume the weight density
Wo = Vo do weight of object
Ww = Vo dw where Ww is weight of equivalent volume of water = 7.37
Wo / Ww = do / dw dividing previous equations
do = 11.2 / 7.37 dw = 1.52 dw
The density of the object is 1.52 that of water
The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3
So the weight density is 14900 N/m^3
An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.
To find the density, the given values are,
Weight in air = 11.20 N
Weight in water = 3.83 N
density of water = 1000 kg/m³
What is meant by Density?According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.
Loss in weight of the object = Weight of object in air - weight of object in water
Loss in weight = 11.20 - 3.83 = 7.37 N
Volume of body x density of water x g = 7.37
Let V be the volume of body
V x 1000 x 9.8 =7.37
V = 7.5× 10⁻⁴ m³
Weight in air = Volume of body x density of body x g
11.20 = 7.5× 10⁻⁴ x d x 9.8
d = 1523 kg/m³.
Thus, the density of body is 1523 kg/m³.
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A cello and an organ are playing together. The organ plays a pitch of C with a frequency of 65.4 Hz in a pipe open at both ends. The cello plays its C string (a string fixed at both ends), and a beat frequency of 2.5 Hz is heard. The cellist loosens the string until no beat frequency is heard. what is the length of the pipe
Answer:
λ/2 = length of pipe
The pipe is open at both ends having wavelength of A-N-A or antinode-node-antinode which is 1/2 wavelength
λ = 331 m/s / (65.4 / s) = 5.06 m wavelength of sound
L = 5.06 m / 2 = 2.53 m length of pipe
