A spring stretches by 21.0 cm when a 135 N object is attached. What is the weight of a fish that would stretch the spring by 31 cm?
Answer:
199.28 N
Explanation:
It is given that,
A spring stretches by 21.0 cm when a 135 N object is attached.
We need to find the weight of a fish that would stretch the spring by 31 cm.
We know that, the force in a spring is given by as per Hooke's law as follows :
F = kx
Where k is spring constant
[tex]\dfrac{F_1}{F_2}=\dfrac{x_1}{x_2}\\\\\dfrac{135}{F_2}=\dfrac{21}{31}\\\\F_2=\dfrac{135\times 31}{21}\\\\F_2=199.28\ N[/tex]
So, the required weight of a fish is 199.28 N.
What is the speed of a jet plane that flies 8100 km in 27 hours (in km/hr)? *
A) 300 km/h
B )8100 km/h
C) 0.0033 km/h
D) 218700 km/h
Answer:
A
Explanation:
if you divide the 8100 km by the 27 hours you will get 300 km per hour
Based on the periodic Table, how many neutrons are most likely in a neutral atom of potassium(K)?
A 20 kg sled is coasting with constant velocity at 5 m/s over perfectly smooth, level ice. It enters a rough stretch of ice 20 m long in which the force of friction is 8 N. With what speed does the sled emerge from the rough stretch
Answer:
vf = 3 m/s
Explanation:
Applying the work-energy theorem, we know that the change in kinetic energy of the object while crossing the rough part of the ice, is equal to the work done by the friction force (as it is the only net force doing work on the sled), as follows:[tex]\Delta K = \frac{1}{2}* m* (v_{f}^{2} -v_{o} ^{2} ) = W_{Ffr} (1)[/tex]
Replacing for the givens, we get:
[tex]\Delta K = \frac{1}{2}* 20kg* (v_{f}^{2} -(5 m/s) ^{2} ) = -20 m * 8 N (2)[/tex]
Rearranging terms, and solving for vf, we get:[tex]v_{f} = \sqrt{-16 (m/s)2 + 25 (m/s)2} = \sqrt{9 (m/s2)} = 3 m/s[/tex]
vf = 3 m/s
what capacitance is required to store an energy of 10kw.h at a potential difference of 1000v?
A Carnot engine receives 250 kJ/s of heat from a heat source reservoir at 525 o C and rejects heat to a heat-sink reservoir at 50 o C. What are the power developed and the heat rejected
Answer:
Explanation:
efficiency of carnot engine = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁
Q₁ = heat absorbed , Q₂ = heat rejected
T₁ = Temperature of source = 525 + 273 = 798 K
T₂ = Temperature of sink = 50 + 273 = 323 K
efficiency = (798 - 323) / 798
= 475 / 798 = .595
(Q₁ - Q₂) / Q₁ = .595
(250 - Q₂) / 250 = .595
250 - Q₂ = 148.75
Q₂ = 101.25 kJ
Heat rejected = 101.25 kJ .
output work = Q₁ - Q₂ = 250 - 101.25 = 148.75 kJ / s
power = 148.75 kJ /s
In a grocery store, you push a 15.5 kg shopping cart with a force of 10.5 N. If the cart starts at rest, how far does it move in 3.00 s?
Answer:
The distance moved by the cart is 3.05 m
Explanation:
Given;
mass of the cart, m = 15.5 kg
applied force, f = 10.5 N
initial velocity, u = 0
time of motion, t = 3s
The final velocity of the cart is given by;
[tex]F = ma\\\\F = m(\frac{v-u}{t})\\\\ m(v-u) = Ft\\\\m(v-0) = Ft\\\\mv = Ft\\\\v = \frac{Ft}{m}\\\\ v = \frac{10.5*3}{15.5}\\\\ v = 2.032 \ m/s[/tex]
The acceleration of the cart is given by;
F = ma
a = F/m
a = 10.5 / 15.5
a = 0.677 m/s²
The horizontal distance moved by the cart is given by;
s = ut + ¹/₂at²
s = 0 + ¹/₂ (0.677)(3)²
s = 3.05 m
Therefore, the distance moved by the cart is 3.05 m
During which stage of sleep, also known as delta sleep, would a person have slow breathing and be difficult to wake.
A.
between stages 1 and 2 of non–rapid-eye-movement sleep
B.
in stage 4 of non–rapid-eye-movement sleep
C.
near the end of stage 3 of rapid-eye-movement sleep
D.
in stage 3 of non–rapid-eye-movement sleep
Answer:
In stage 3 so the answer is D
Answer:
It's D. in stage 3 of non-rapid-eye-movement sleep
Explanation:
**PLATO** "NREM-3 is when a person is in deep sleep and it lasts about 30 minutes. The person experiences slow breathing and pulse rates and limp muscles, and is difficult to wake. This stage is sometimes called delta sleep, because the brain produces slow delta waves."
Why do objects have specific colors?
Answer:
The color of an object is the wavelength of light that it reflects.
An astronaut on the surface of Mars standing on the edge of a 73.3 meter tall cliff throws a golf ball
down with an initial speed of 4.5 m/s. The ball hits the ground 5.2 seconds later. Find the
acceleration due to gravity on Mars.
Answer:
3.69m/s²
Explanation:
Using the formula;
S = ut + 1/2gt²
Where;
S = distance (m)
u = initial velocity (m/s)
t = time (s)
g = accelerate due to gravity (m/s)
Based on the information provided, S = 73.3m, u = 4.5m/s, t = 5.2s, g= ?
73.3 = (4.5 × 5.2) + 1/2 (g × 5.2²)
73.3 = 23.4 + 1/2 (27.04g)
73.3 = 23.4 + 13.52g
73.3 - 23.4 = 13.52g
49.9 = 13.52g
g = 49.9/13.52
g = 3.6908
g = 3.69m/s²
a model rocket of mass 0.5 kg is launched straight up. At an altitude of 140 m, when its speed is 90 m/s, it spontaneously blows apart into two equal pieces. One piece continues upward at a speed of 10 m/s. How fast is the other piece moving
Answer:
The other piece is moving at 170 m/s.
Explanation:
To find the speed of the second piece can be found by conservation of linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
Where:
m₁: is the rocket's mass = 0.5 kg
m₂ = m₃: is the mass of the two equal pieces = 0.25 kg
v₁: is the rocket speed = 90 m/s
v₂: is the speed of the first piece = 10 m/s
v₃: is the speed of the second piece =?
[tex] 0.5*90 = 0.25(10 + v_{3}) [/tex]
[tex] v_{3} = \frac{45}{0.25} - 10 = 170 m/s [/tex]
Therefore, the other piece is moving at 170 m/s.
I hope it helps you!
Consider a rocket (initial mass m o) accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases p eventually begins to decrease. For what value of m is p maximum
Answer:
m = m_o/e
Explanation:
The rocket equation is given as;
m(dv/dt) = -v_ex(dm/dt)
v_ex is the exhaust velocity
Now, formula for momentum is;
p = mv
Differentiating with respect to time, we have;
dp/dt = m'v + mv'
Where;
m' is mass rate
v' is rate of change in velocity
Since we are dealing with exhaust velocity and momentum(p) is Maximum when v = v_ex then mv' can also be written as: -m'v_ex.
Thus;
dp/dt = m'v - m'v_ex
dp/dt = m'(v - v_ex)
Now, from the rocket equation we dt will cancel out to give;
-dv/v_ex = dm/m
Integrating both sides;
-∫dv/v_ex = ∫dm/m
This gives;
-v/v_ex = In(m/m_o)
Since momentum(p) is Maximum when v = v_ex
Thus;
-v/v = In(m/m_o)
-1 = In(m/m_o)
e^(-1) = m/m_o
m = m_o(e^(-1))
m = m_o/e
A train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s. What is the change in velocity of the train
The change in velocity of the train is 120 m/sec.
What is Velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity. Unit of velocity is m/sec.
Given in the question train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s,
We see that the change in direction of the train is 180 degree,
the net change in velocity is given as,
v = 60 m/s + 60 m/s
v = 120 m/s
A train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s the change in velocity of the train is 120 m/sec.
To learn more about velocity refer to the link:
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5n right 5n left I don’t know this please help!
Answer:
this is my exception I am not sure but I think its help full 5Nr+5NL so it will be 0N which is equilibrium
Answer the following: What are the units of mass and the units of weight for the metric system? What are the units of mass and the units of weight for the English system?
Answer:
Explanation:
Units are measure of a quantity. There are two types of units,
The fundamental units
The derived units
The fundamental units are units that are independent i.e they can stand alone while derived units are derived from fundamental units (They are dependent)
The fundamental units are the units of fundamental quantities (Length, mass and time)
Length is measured in metres (m)
Mass is measured in kilograms (kg)
Time is measured in seconds(s)
This are called the standard units where other units are derived.
Other units of mass are grams
Weight is the product of mass and acceleration due to gravity
Weight = mass×acc due to gravity
W = mg
Using units
W = kg × m.s²
Since acceleration is measured in m/s²
W = kgm/s²
Hence the metric unit of weight is kgm/s² or Newtons
If the wire is lowered farther from the compass, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection?
Answer:
The new angle of deflection north pole becomes smaller.
Explanation: The angle of deflection is the angle formed when an object changes course from its original course of direction or target.
The angle of deflection of a particular particle is directly proportional to its charge to mass ratio as it passed through an electric field.
As the wire is continously being lowered farther from the compass,the angle of deflection of the north pole ontinues to become smaller when compared to its initial deflection.
An oxygen-16 ion with a mass of perpendicular to a 1.20-T magnetic field, which makes it move in a circular arc with a 0.231-m radius.
Required:
a. What positive charge is on the ion?
b. What is the ratio of this charge to the charge of an electron?
c. Discuss why the ratio found in (b) should be an integer.
Answer:
a) 4.8*10^-19 C
b) 3
Explanation:
We know that
F = ma
Also, recall that a = v²/r and F = qvB
Substituting this in the first equation, we have
qvB = mv²/r, since we're looking for q, we make it the subject of the formula and solve for it.
q = mv²/rvB
q = mv/rB
q = [2.66*10^-26 * (5*10^6)] / 0.231 * 1.2
q = 1.33*10^-19 / 0.2772
q = 4.8*10^-19 C
Ratio between the charge and the electron is q/c =
4.8*10^-19 / 1.6*10^-19 = 3
Therefore, the ratio between the charge and electron is 3
What is the total energy transported per hour along a narrow cylindrical laser beam 1.80 mm in diameter, whose B-field has an rms strength of
Am assuming the B-field has an rms strength of 1.10×10−10T which was omitted
Answer: E=1.686 x10⁻⁸J/hr
Explanation:
Intensity for electromagnetic wave is given as
I= cε₀(Emax)²....... equation 1
Also
B=Emax/c ....... equation 2
puting the value of B in equation 2 in equation 1 becomes
I= cε₀(Emax)²
I= c³ε₀B²
where
c = speed of light = 3 X 10⁸m/s
ε₀=permittivity of free space = 8.85x 10-12 F m-1
B=1.10×10−10T
I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²
I=2.891 X 10⁻⁶ W/m²
Energy per unit hour
E= IAt
= 2.891 X 10⁻⁶ W/m² x πr²x 3600
2.891 X 10⁻⁶ W/m² x 3.142 x ( 1.80 x 10⁻³/2) x(3600/1hr)
E=1.686 x10⁻⁸J/hr
The total energy transported per hour will be 6.62 × 10⁻⁹J
Calculation of energy transported:The Intensity for an electromagnetic wave is given by:
I = cε₀E₀²
where c is the speed of light, and
E₀ is the maximum value of the electric vector
Also
B = E/c
where B is the magnetic field
Therefore:
I= cε₀E₀²
I= c³ε₀B²
Given that the magnetic field is:
B = 1.10×10⁻¹⁰ T
I= (3 X 10⁸m/s)³ X (8.85 X 10⁻12F/m) x(1.10×10−10T)²
I=2.891 X 10⁻⁶ W/m²
Energy transported per unit hour
E= IAt
where A is the area
and t is the time = 1hr = 3600s
E = 2.891 × 10⁻⁶ × πr² × 3600
E = 2.891 X 10⁻⁶ × 3.14 × ( 1.80 x 10⁻³/4)² x (3600)
E = 6.62 × 10⁻⁹J
Learn more about electromagnetic waves:
https://brainly.com/question/13803241?referrer=searchResults
A cart with a mass of 55 kg is pushed with a force of 220 N. What is the acceleration?
Answer:
4m/s^2
Explanation:
According to newton's second law of motion:
F= ma
a= F/m
F =220N
m =55kg
a= 220/55
= 4m/s^2
A ship maneuvers to within 2.47×10^3 m of an island’s 1.77 × 10^3 m high mountain peak and fires a projectile at an enemy ship 6.02 × 10^2 m on the other side of the peak, as illustrated. The ship shoots the projectile with an initial velocity of 2.55×10^2 m/s at an angle of 72.4. The acceleration of gravity is 9.81 m/s 2. How close (vertically) does the projectile
come to the peak?
I've attached a sketch of what I interpret the situation to be. It seems to me that you have to find the height of the projectile above the mountain's peak, assuming it actually does hit the ship on the other side. (Note that the sketch is not drawn to-scale. I don't mean to suggest that the projectile reaches its maximum height directly above the mountain.)
Compute the horizontal and vertical components of the velocity:
[tex]v_{i,x}[/tex] = (2.55 x 10² m/s) cos(72.4º) ≈ 77.1 m/s
[tex]v_{i,y}[/tex] = (2.55 x 10² m/s) sin(72.4º) ≈ 243 m/s
The projectile's horizontal and vertical positions at time t from its launching point are
x = [tex]v_{i,x}[/tex] t
y = [tex]v_{i,y}[/tex] t - 1/2 (9.81 m/s²) t²
Find the time it takes for the projectile to travel the distance from the first ship to the mountain's peak:
2.47 x 10³ m = (77.1 m/s) t
t = (2.47 x 10³ m) / (77.1 m/s)
t ≈ 32.0 s
Find the vertical position at this time:
y = (243 m/s) (32.0 s) - 1/2 (9.81 m/s²) (32.0 s)²
y ≈ 2750 m = 2.75 x 10³ m
Take the difference of this height and the height of the mountain:
2.75 x 10³ m - 1.77 x 10³ m = 0.980 x 10³ m = 9.80 x 10² m
In riot control, the riot squad uses a water hose that shoots water at a rate of 5 m/s and volume of 30 L/s. What is the average force exerted on a person assuming that the water splashes sideways in all directions? The density of water is 1000 kg/m3.
Answer:
The average force exerted on a person assuming that the water splashes sideways in all directions is 150 newtons.
Explanation:
From Fluid Mechanics and Newton's Laws we know that water stream is an example of a variable mass system at constant speed and the magnitude of the net force done by the water splash ([tex]F[/tex]), measured in newtons, is given by the following expression:
[tex]F = \rho\cdot \frac{dV}{dt}\cdot v[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of water, measured in kilograms per cubic meter.
[tex]\frac{dV}{dt}[/tex] - Volume flow rate, measured in cubic meters per second.
[tex]v[/tex] - Flow velocity, measured in meters per second.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\frac{dV}{dt} = 0.03\,\frac{m^{3}}{s}[/tex] and [tex]v = 5\,\frac{m}{s}[/tex], the average force exerted on a person is:
[tex]F = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(0.03\,\frac{m^{3}}{s} \right)\cdot \left(5\,\frac{m}{s} \right)[/tex]
[tex]F = 150\,N[/tex]
The average force exerted on a person assuming that the water splashes sideways in all directions is 150 newtons.
3. What would be the mass of a truck if it is accelerating at a rate of 5 m/s and hits a parked car with a
force of 14,000 N?
I
Answer:
We are given both the Force applied by the truck and the acceleration of the truck:
Force applied (F) = 14,000 N
Acceleration (a)= 5 m/s²
Solving for the mass of the truck:
From Newton's second law of motion,
F = ma (where F is the force applied , m is the mass of the object and a is the acceleration of the object)
replacing the variables:
14000 = (m)* 5
m = 14000/5
m = 2800
Therefore, the mass of the truck is 2800 kg
Answer:
m = 2800 N
Explanation:
14000 = (m)* 5
m = 14000/5
F= 14,000 N
a= 5 m/s^2
need help with momentum!
Answer:
The total kinetic energy must be greater than the momentum of either ball
Explanation:
In order to solve this problem, we must clarify the following conditions, the balls are in motion that is, there is kinetic energy, the kinetic energy is a scalar magnitude, therefore this energy is greater than zero.
But however momentum is equal to zero, momentum is a vector quantity, that is, these velocities have a direction, momentum is defined as the product of mass by Velocity.
P = m*v (momentum lineal)
where:
P = lineal momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
Since the masses of the balls are equal, we have:
m1 = m2
therefore:
m1*v1 + (m2*v2) = 0 (total momentum)
For the sum to be equal to zero, besides the equivalation of the masses, the velocities must also be equal, but the velocities must be of opposite sign, to meet equality to zero.
m1*v1 - m2*v2 = 0
And the kinetic energy of a body is defined by means of the following equation
Ek = 0.5*m*v²
Since the magnitude of the velocities is greater than zero, so that there is movement, we can say that the kinetic energy is much greater than zero, as well as greater than the total linear momentum
A 872 g wooden block is initially at rest on a rough horizontal surface when a 15.6 g bullet is fired horizontally into (but does not go through) it. After the impact, the block-bullet combination slides 6.50 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet immediately before impact.
Answer:
The speed of the bullet before the impact is 556.5 m/s
Explanation:
Given;
mass of the wooden block, m₁ = 872 g = 0.872 kg
initial velocity of the wooden block, u = 0
mass of the bullet, m₂ = 15.6 g = 0.0156 kg
distance traveled by the block-bullet system after impact, d = 6.5 m
coefficient of kinetic friction, μ = 0.75
Work done by friction on the system after impact is given by;
W = (m₁ + m₂)μg x d
kinetic energy of the system after impact is given by;
K.E = ¹/₂(m₁ + m₂)v²
Apply work energy theorem;
(m₁ + m₂)μg x d = ¹/₂(m₁ + m₂)v²
μg x d = ¹/₂v²
v² = 2μg x d
v = √(2μg x d)
v = √(2 x 0.75 x 9.8 x 6.5)
v = 9.78 m/s
The speed of the bullet before impact is calculated by applying principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.872(0) + 0.0156u₂ = 9.78(0.872 + 0.0156)
0 + 0.0156u₂ = 8.681
u₂ = 8.681 / 0.0156
u₂ = 556.5 m/s
Therefore, the speed of the bullet before the impact is 556.5 m/s
A 95 N net force is applied to an ice block with a mass of 24 kg. Find the acceleration of the block if it moves on a smooth horizontal surface
Answer:
The answer is 3.96 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
f is the force
m is the mass
From the question we have
[tex]a = \frac{95}{24} \\ = 3.95833333333...[/tex]
We have the final answer as
3.96 m/s²Hope this helps you
PLEASE HELP
You are walking to school one day and can't wait to give Mr. DeBaz the homework you finished. As you are walking down the street, a zombie comes out from behind the Dunkin' Donuts. He looks at you, growls, and starts coming after you. You are being chased by a zombie and you run 2 m/s. You are trying to get to the school, for safety, and it is 250 m away. How long will it take you to get there?
Given that the tensile strength of aluminum foil is 311 megapascals, its thickness is approximately 15.0 micrometers, and a roll of household aluminum foil is 30.0 centimeters wide, how much force F is needed to pull off a sheet to use
Answer:
Explanation:
cross sectional area = width x thickness
A = 30 x 10⁻² x 15 x 10⁻⁶ = 450 x 10⁻⁸ m
tensile strength F / A = 311 x 10⁶ N / m²
F = A x 311 x 10⁶
= 450 x 10⁻⁸ x 311 x 10⁶
= 139950 x 10⁻²
= 1399.5 N .
A car accelerates from a standstill to 60 km/hr in 10.0 seconds, what is its acceleration?
a= vf - vi/t
a= 60-0/10
a= 60/10
a= 6 m/s^2
Answer:
6 km/hr/s
Explanation:
A marble rolls off a ledge with an initial velocity of (1.125m/s, 0o). The marble lands 0.45m from the base of the ledge. Calculate the magnitude of the velocity, in m/s, of the marble just before it lands.
I don't want the answer but can someone please describe to me how I'm supposed to solve this and the variables I need.
Answer:
3.2m/s
Explanation:
Given parameters:
Initial velocity = 1.125m/s
Height of fall = 0.45m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to look at our given parameters and compare to the appropriate motion equation.
we have been given; u (initial velocity) and height of fall (h)
Since this body falls under acceleration due to gravity, g = 9.8m/s²
Now, we use;
V² = U² + 2gh
where V is the final velocity
Input the parameters and solve;
V² = 1.125² + (2 x 9.8 x 0.45)
V = 3.2m/s
Two pendulum bobs of unequal mass are suspended from the same fixed point by strings of equal length. The lighter bob is drawn aside and then released so that it collides with the other bob on reaching the vertical position. The collision is elastic. What quantities are conserved in the collision?
Answer:
"Both kinetic energy and angular moment of the system" is the correct solution.
Explanation:
The kinetic energy is:
[tex]KE=\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}v_{2}^2[/tex]
Its conversion is:
⇒ [tex](KE)_{1}=(KE)_{f}[/tex]
Angular momentum is:
[tex]L=I_{1}\omega_{1}+I_{2}\omega_{2}[/tex]
Its conversion is:
⇒ [tex]L_{i}=L_{f}[/tex]
So that both "KE" and "L" of the system conserved.
