Answer:
a = 4 [m/s^2]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration.
F = m*a
where:
F = force = 40 [N]
m = mass = 10 [kg]
a = 40/10
a = 4 [m/s^2]
I don't quite understand. Can you help please?
d2 = 20 m, d1= 50m
What is the magnitude of the resultant?
Answer:
53.85m
Explanation:
If we are assuming that d1 is one side of a triangle, and d2 is the other, and we are looking for the magnitude, which is essentially the hypotenuse, we use Pythagorean Theorem. a^2+b^2=c^2. 20^2 + 50^2 = 2900.
The square root of 2900 is 53.85164. Therefore, that is your hypotenuse.
HELP PLSSSSSSSSSjjdndnsnsj
Answer:
i feel like 3 not too sure tho
Find the magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B.
Answer:
[tex]B=1.61\times 10^{-4}\ T[/tex]
Explanation:
The attached figure shows the path followed by an electron in the semicircular path from A to B.
Velocity of the electron is, [tex]v=1.42\times 10^6\ m/s[/tex]
It can be seen from the figure that the radius of thenpath, r = 5 cm or 0.05 m
The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,
[tex]Bqv=\dfrac{mv^2}{r}[/tex]
B is the magnitude of the magnetic field
[tex]B=\dfrac{mv}{rq}\\\\\text{Putting all the values}\\\\B=\dfrac{9.1\times 10^{-31}\times 1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}\\\\B=1.61\times 10^{-4}\ T[/tex]
So, the magnitude of the magnetic field is [tex]1.61\times 10^{-4}\ T[/tex].
The magnitude of the magnetic field that will cause the electron to follow the semicircular path from A to B will be [tex]B=1.61\times 10^-{4]\ T[/tex]
What is magnetic field?The magnetic field is defined as when the current passes through the wire, then the magnetic field is generated around the wire in a circular pattern.
The attached figure shows the path followed by an electron in the semicircular path from A to B.
The velocity of the electron is, [tex]v=1.42\times 10^6\ \frac{m}{s}[/tex]
It can be seen from the figure that the radius of then path, r = 5 cm or 0.05 m
The magnetic force acting on the electron is balanced by the centripetal force acting on it. It means,
[tex]Bqv=\dfrac{mv^2}{r}[/tex]
B is the magnitude of the magnetic field
[tex]B=\dfrac{mv}{rq}[/tex]
[tex]B=\dfrac{9.1\times 10^{-31}\times1.42\times 10^6}{0.05\times 1.6\times 10^{-19}}[/tex]
[tex]B=1.61\times 10^{-4}\ T[/tex]
So, the magnitude of the magnetic field is [tex]B=1.61\times 10^-{4]\ T[/tex]
To know more about the magnetic field, follow
https://brainly.com/question/26257705
Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 2.10 cm from the axis to equal 5.00×105 g (where g is the free-fall acceleration)
Answer:
ω = 15275.25 rad/s
Explanation:
Given that,
Radial acceleration of an ultracentrifuge is, [tex]a=5\times 10^5g[/tex]
Distance from the axis, r = 2.1 cm = 0.021 m
g is the free-fall acceleration such that g = 9.8 m/s²
We need to find the angular speed of an ultracentrifuge. The formula that is used to find the angular speed is given by formula as follows :
[tex]a=r\omega^2[/tex]
Putting all the values,
[tex]\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5\times 10^5\times 9.8}{0.021}} \\\\\omega=15275.25\ rad/s[/tex]
So, the required angular speed, ω, of an ultracentrifuge is 15275.25 rad/s.
What effect docs temperature have on the dissolution rate in sugar water?
mechanical energy defintion
Answer: Mechanical energy is the energy that is possessed by an object due to its motion or due to its position.
(The energy acquired by the objects upon which work is done)
the rotational speed of earth is similar to?
PLS URGENT!!
A radio station sending out a radio wave of frequency 100.5MHz at velocity of 3×10⁸ms⁻¹. At what wavelength is the radio station broadcasting
Which best illustrates the electromagnetic force in action?
-a football being kicked
-leaves falling from tree
-flashlight
-neutron beta particle and proton
Answer:
neutron beta particle and proton (last option in the list)
Explanation:
The neutron beta particle and proton inside a neutron is a clear example of a negative particle (beta particle) and a positive particle (proton) experiencing electromagnetic force (attraction between positive and negative charges) at a very short distance.
Answer:
I'm pretty sure it's the flashlight because electromagnetic force produces electricity.
If this piece of abductin is 3.1 mm thick and has a cross-sectional area of 0.49 cm2 , how much potential energy does it store when compressed 1.5 mm ?
Complete question:
A scallop forces open its shell with an elastic material called abductin, whose Young's modulus is about 2.0×10⁶ N/m2 .
If this piece of abductin is 3.1 mm thick and has a cross-sectional area of 0.49 cm2 , how much potential energy does it store when compressed 1.5 mm ?
Answer:
The elastic potential energy of the material is 0.036 J
Explanation:
Given;
Young's modulus, E = 2.0×10⁶ N/m²
Thickness of the abductin, l = 3.1 mm = 0.0031 m
compression of the abductin, x = 1.5 mm = 0.0015 m
area, A = 0.49 cm² = 0.49 x 10⁻⁴ m²
Young's modulus for elastic material is given by;
[tex]E = \frac{stress}{strain} = \frac{Fl}{Ax} \\\\ E = \frac{F}{x}*\frac{l}{A}\\\\ E = k*\frac{l}{A}\\\\k = \frac{AE}{l}\\\\k = \frac{(0.49 x10^{-4})(2*10^6)}{0.0031}\\\\ k = 31,612.9 \ N/m[/tex]
The elastic potential energy of the material is given by;
U = ¹/₂kx²
U = ¹/₂(31,612.9)(0.0015)²
U = 0.036 J
Therefore, the elastic potential energy of the material is 0.036 J
If the forces on an object are balanced, the object will do what?
Stokes’ law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density 7.8 \times 10^3~\text{kg/m}^37.8×10 3 kg/m 3 and diameter 3.0~\text{mm}3.0 mm) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.
Answer:
The viscosity is [tex]\eta = 0.76243 \ kg/ m \cdot s [/tex]
Explanation:
From the question we are told that
The density is [tex]\rho = 7.80 *10^{3} \ kg/m^3[/tex]
The diameter is [tex]d = 3.0 \ mm =0.003 \ m[/tex]
The time taken is [tex]t = 12 \ s[/tex]
The distance covered is [tex]d = 0.60 \ m[/tex]
Generally the velocity of the ball is
[tex]v = \frac{d}{t}[/tex]
=> [tex]v = \frac{0.60}{12}[/tex]
=> [tex]v = 0.05 \ m/s [/tex]
Generally the mass of the steel ball is
[tex]m = \rho * V[/tex]
Here V is the volume and this is mathematically represented as
[tex]V = \frac{4}{3} * \pi * [\frac{d}{2} ]^3[/tex]
=> [tex]V = \frac{4}{3} * 3.142 * [\frac{0.003}{2} ]^3[/tex]
=> [tex]V = 1.414 *10^{-8} \ m^3[/tex]
So
[tex]m = 7.80 *10^{3} * 1.414 *10^{-8}[/tex]
[tex]m = 0.00011 \ kg [/tex]
Generally the viscosity is mathematically represented as
[tex]\eta = \frac{m * g}{6\pi * r * v }[/tex]
Here r is the radius represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.003}{2}[/tex]
[tex]r = 0.0015 \ m [/tex]
So
[tex]\eta = \frac{0.00011 * 9.8}{6 * 3.142 * 0.0015 * 0.05 }[/tex]
=> [tex]\eta = 0.76243 \ kg/ m \cdot s [/tex]
A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.
Answer:
The magnitude of the friction force exerted on the box is 2.614 newtons.
Explanation:
Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:
[tex]\Sigma F = -f = m\cdot a[/tex] (Eq. 1)
Where:
[tex]f[/tex] - Kinetic friction force, measured in newtons.
[tex]m[/tex] - Mass of the box, measured in kilograms.
[tex]a[/tex] - Acceleration experimented by the box, measured in meters per square second.
By applying definitions of weight ([tex]W = m\cdot g[/tex]) and uniform accelerated motion ([tex]v = v_{o}+a\cdot t[/tex]), we expand the previous expression:
[tex]-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex]
And the magnitude of the friction force exerted on the box is calculated by this formula:
[tex]f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)[/tex] (Eq. 1b)
Where:
[tex]W[/tex] - Weight, measured in newtons.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]v[/tex] - Final speed, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]W = 52.4\,N[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 1.37\,\frac{m}{s}[/tex], [tex]v = 0\,\frac{m}{s}[/tex] and [tex]t = 2.8\,s[/tex], the magnitud of the kinetic friction force exerted on the box is:
[tex]f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)[/tex]
[tex]f = 2.614\,N[/tex]
The magnitude of the friction force exerted on the box is 2.614 newtons.
A parallel plate capacitor contains a positively charged plate on the left, and a negatively charged plate on the right. An electron in between the plates is moving to the right. Which statement is true? Group of answer choices The potential energy of the electron is decreasing and it is moving to a region having a higher potential The potential energy of the electron is decreasing and it is moving to a region having a lower potential. The potential energy of the electron is increasing and it is moving to a region having a lower potential The potential energy of the electron is increasing and it is moving to a region having a higher potential.
Complete Question
A parallel plate capacitor contains a positively charged plate on the left, and a negatively charged plate on the right. An electron in between the plates is moving to the right. Which statement is true?
Group of answer choices
a The potential energy of the electron is decreasing and it is moving to a region having a higher potential
b The potential energy of the electron is decreasing and it is moving to a region having a lower potential.
c The potential energy of the electron is increasing and it is moving to a region having a lower potential
d The potential energy of the electron is increasing and it is moving to a region having a higher potential.
Answer:
The correct option is C
Explanation:
From the question we are told that
An electron in between the plates is moving to the right
Generally the potential energy of the electron is mathematically represented as
[tex]PE = e * V[/tex]
Here e is the charge on the electron and V is the electric potential of the electron
Generally the left side with the positive charge has a higher potential than the right side with the negative charge
Now when the potential energy of the electron increases it will move toward the plate with the lower potential which is the left plate
How much is the weight of a 1 kg mass at the pole and the equator of the earth
Given parameters:
Mass given = 1kg
Unknown:
Weight of the body at pole and equator = ?
Solution:
Both locations are on the surface of the earth. Generally, we take 9.8m/s² as the acceleration due to gravity on the earth.
Weight = mass x acceleration due to gravity
But little disparity occurs in the value of acceleration due to gravity from the pole to equator. This is due to equatorial bulge.
At the equator , 9.780 m/s²
pole 9.832 m/s²
Weight at equator = 1 x 9.780 = 9.78N
Weight at the pole = 1 x 9.832 = 9.832N
Please help I don’t know how to do this
Answer:
3.176 hoursExplanation:
given:
distance = 270 km
speed = 85 km/h
find:
how long does it take to get into his audition in hours?
solution:
velocity = distance / time
85 km/h = 270 km
t
85 (t) = 270
t = 270 / 85
t = 3.176 hours
A force of 50 N acts upon a 10 kg block. Calculate the acceleration of the object.
Explain why a golf ball that is hit on the moon travels much farther than a golf ball that is hit with the same force on Earth.
Answer:
the reason why is that if it is hit with the same force the moon has a lot less gravity than the earth dose so the golf ball will hit the ground on earth first before the golf ball on moon dose
What happens to the temperature of a substance while it is changing state?
Answer:
its temperature stays constant
Explanation:
explain what happent to the pressure exerted by an object when the area over which it is exerted:
a) increase
b) decrease
An electromagnet crane is carrying the electromagnet with the help of the three cables. But the electromagnet is not stable because of the wind. What is the reason behind the cause
Answer:
The reason behind the given instance is summarized below.
Explanation:
When another electromagnet shifts because of the wind, it induces a simulated area that acts on either the electromagnet as an effect or force, considered as the Lorentz force. This force thus renders the resultant force equivalent to a non-zero value mostly on the electromagnet. The balance is not protected by this power. When there is no breeze, just the three connectors sustain equilibrium.A .25 kilogram baseball is thrown upwards with a speed of 30 meters per second. Neglecting friction the maximum height reached by the baseball is approximately
Answer: 46 meters
Explanation:
The maximum height reached by the base ball is 45.9 meters.
To find the maximum height the given values are,
Mass = 25 Kg
speed v = 30 meters per second.
What is the height of baseball reached?As the ball goes up, it acts due to the gravity.
The gravity slows an upward-moving object,
Acceleration due to gravity = 9.8 m/s every second,
the baseball reaches its greatest height in (30/9.8) = 3.06 seconds.
At that instant, its speed is zero.
The baseball's average speed from toss to peak is
(1/2) (30 + 0) = 15 m/s .
The baseball average speed = 15 m/s
And the time for greatest height = 3.06 seconds.
Substituting the given values,
The ball rises can be calculated as
(15 x 3.06) = 45.9 meters.
Thus, the baseball reached the maximum height approximately is 45.9 meters.
Learn more about the height due to gravity,
https://brainly.com/question/1977173
#SPJ6
An object is traveling with a velocity of 3.05 m/s. The object then accelerates at 2.82 m/s-over a
displacement of 18.4 m. What is its final velocity?
Answer:
Final velocity (v) = 5.148 m/s (Approx)
Explanation:
Given:
Initial velocity (u) = 3.05 m/s
Acceleration (a) = 2.82 m/s²
Displacement (s) = 18.4 m
Find:
Final velocity (v)
Computation:
v² = u² + 2as
v² = (3.05)² + 2(2.82)(3.05)
v² = 9.3025 + 17.202
v² = 26.5045
v = 5.148 m/s (Approx)
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 29.9 m/s2 with a beam of length 5.33 m , what rotation frequency is required
Answer:
Rotation frequency is 0.377 hertz.
Explanation:
After a careful reading of statement, we need to apply the concept of radial acceleration due to uniform circular motion, whose formula is:
[tex]a_{r} = \omega^{2}\cdot L[/tex] (Eq. 1)
Where:
[tex]a_{r}[/tex] - Radial acceleration, measured in meters per square second.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]L[/tex] - Length of the beam, measured in meters.
Now we clear the angular velocity within:
[tex]\omega = \sqrt{\frac{a_{r}}{L} }[/tex]
If [tex]a_{r} = 29.9\,\frac{m}{s^{2}}[/tex] and [tex]L = 5.33\,m[/tex], the angular velocity is:
[tex]\omega = \sqrt{\frac{29.9\,\frac{m}{s^{2}} }{5.33\,m} }[/tex]
[tex]\omega \approx 2.368\,\frac{rad}{s}[/tex]
The frequency is the number of revolutions done by device per second and can be found by using this expression:
[tex]f = \frac{\omega}{2\pi}[/tex] (Eq. 2)
Where [tex]f[/tex] is the frequency, measured in hertz.
If we know that [tex]\omega \approx 2.368\,\frac{rad}{s}[/tex], then rotation frequency is:
[tex]f = \frac{2.368\,\frac{rad}{s} }{2\pi}[/tex]
[tex]f = 0.377\,hz[/tex]
Rotation frequency is 0.377 hertz.
What is the ostrich’s average acceleration from 9.0 to 18s
Answer:
Explanation:
10.00 5
What was Bear B's average speed in m/min? (No units required for answer. Type your answer.)
Answer:
43
Explanation:
you divide 2600 by 60 and your answer is 43.3333333
I just took the quiz and got it right!
derive ideal gas equation for n mole of gas.
Answer:
Explanation:
Ideal gas equation- The volume (V) occupied by the n moles of any gas has pressure(P) and temperature (T) Kelvin,
the relationship for these variables PV=nRT where R gas constant is called the ideal gas law
Derivation of the Ideal Gas Equation
Let us consider the pressure exerted by the gas to be ‘p,’
The volume of the gas be – ‘v’
Temperature be – T
n – be the number of moles of gas
Universal gas constant – R
According to Boyle’s Law,
it constant n & T, the volume bears an inverse relation with the pressure exerted by a gas.
i.e. v∝1p ………………………………(i)
According to Charles’ Law,
When p & n are constant, the volume of a gas bears a direct relation with the Temperature.
i.e. v∝T ………………………………(ii)
According to Avogadro’s Law,
When p & T are constant, then the volume of a gas bears a direct relation with the number of moles of gas.
i.e. v∝n ………………………………(iii)
Combining all the three equations, we have-
v∝nTp
or pv=nRT
where R is the Universal gas constant, which has a value of 8.314 J/mol-K
If you hold a piece of copper wire in an open flame, which of the following will most likely happen?
The heat will travel through the wire to your hand.
The light will change into electricity.
The flame will absorb heat from the wire.
The ſame will get brighter.
Answer:
The heat will travel through the wire to your hand.
Explanation:
Thats the one that makes the most sense.
A Coolie raises a box of mass 20 kg to a height of 1.5m. If the force of gravity on 1 kg is 10N. The work done by the Coolie is
