Answer:
The answer is option AExplanation:
To find the mass of an object given it's acceleration and the force acting on it we use the formula
[tex]mass = \frac{force}{acceleration} \\ [/tex]
From the question
force = 500 N
acceleration = 5.0 m/s²
We have
[tex]mass = \frac{500}{5} \\ [/tex]
We have the final answer as
100 kgHope this helps you
Do 2. A cyclist starts from rest and accelerates along a straight path to a speed of
12.15 m/s in a time of 4.5 seconds.
What is the cyclist's acceleration to the nearest tenth?
O
54.7 m/s2
o
24.3 m/s?
2.7 m/s2
De
3.4 m/s?
The cyclist's acceleration to the nearest tenth is: C. 2.7 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0 m/s (since the cyclist starts from rest).Final velocity = 12.5 m/s.Time = 4.5 seconds.To find the cyclist's acceleration to the nearest tenth:
Acceleration is calculated by subtracting the initial velocity from the final velocity and dividing by the time.
Mathematically, acceleration is given by the formula;
[tex]A = \frac{V - U}{t}[/tex]
Where:
A is the acceleration.V is the final velocity.U is the initial velocity.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]A = \frac{12.5\; - \;0}{4.5} \\\\A = \frac{12.5}{4.5}[/tex]
Acceleration, A = 2.7 [tex]m/s^2[/tex]
Therefore, the cyclist's acceleration to the nearest tenth is 2.7 [tex]m/s^2[/tex].
Read more: https://brainly.com/question/8898885
URGENT!! A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
momentum of -2.80 kg m/s and a final momentum of -1.97 kg-m/s. The ball
has an initial momentum of 0.02565 kg-m/s. If you assume the collision is
elastic, what is the final velocity of the ball?
O A. -83.26 m/s
O B. -0.80 m/s
O C. -14.11 m/s
O D. -4.75 m/s
Answer:
-14.11
Explanation:
I don't quite understand. Can you help please?
5
2. When you keep the mass the same and decrease the unbalanced force how does the acceleration change?
Answer:
The acceleration will decrease due to less force acting on the object.
Explanation:
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
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
Types of muscle tissue includes
Skeletal muscle
Cardiac muscle
Smooth muscle
All of the above
Answer:
D. all of the above
Explanation:
hope it helps
Two cars traveling in the same direction pass you at exactly the same time . The car that is going faster
moves farther in the same amount of time.
Explanation:
based on my answer......
we supposed to assume that both cars are equal except for speed. So they would have the same mass.....
Anyway, the question doesn't tell you anything about weight or size, so you can't tell from the information given which has more mass.
But the very definition of 'faster' is 'moves farther in the same amount of time'.
Hope this helps
Answer:
They both moves at the same speed
Explanation:
hope it works
HELP PLSSSSSSSSSjjdndnsnsj
Answer:
i feel like 3 not too sure tho
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
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.
Sedimentary rock turns into magnum through which process
I WILL MARK YOU AS BRAINLIEST IF RIGHT
In order to climb up a mountainside, a train needs to start at a speed of 40 mph. The speed limit on the track, however, is 25 mph. How much time does the train need to get up to 40 mph if it can accelerate 4 mph per minute ?
Let's see what to do buddy.....
_________________________________
[tex]Acceleration = 4 \: \: \frac{mile}{ h } p( min ) [/tex]
Thus ;
The speed increases 4 units per minute.
The train has a speed of 25 mph now.
After a minute it has a speed of 29 mph.
After a minute it has a speed of 33 mph.
After a minute it has a speed of 37 mph.
Look ; 4 units per minute means 4 units per 60 seconds, thus the speed increases 1 unit per 15 seconds .
If the speed wants to increasing from 37 to 40 , needs 3 × 15 seconds = 45 sec.
So we have : 1 min + 1 min + 1 min + 45 sec .
3 minutes and 45 seconds or 225 seconds need to the speed increasing from 25 mph to 40 mph.
_________________________________
And we're done.....♥️♥️♥️♥️♥️
leaves uses_,_and_to make food for the plant
light, water, carbon dioxide
Explanation:
c02 , h20 and light
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?
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.
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,
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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.
explain what happent to the pressure exerted by an object when the area over which it is exerted:
a) increase
b) decrease
Where is the magnetic south pole compared to the geographical north pole?
Currently, the magnetic south pole lies about ten degrees distant from the geographic north pole, and sits in the Arctic Ocean north of Alaska. The north end on a compass therefore currently points roughly towards Alaska and not exactly towards geographic north.
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
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
object above the surface of the earth, in miles, for the object to register a weight of 125
pounds.
Answer:
The height of the object is 5007.4 miles.
Explanation:
Given that,
Weight of object = 200 lb
We need to calculate the value of [tex]Gmm_{e}[/tex]
Using formula of gravitational force
[tex]F=\dfrac{Gmm_{e}}{r^2}[/tex]
Put the value into the formula
[tex]200=\dfrac{Gmm_{e}}{(3958.756)^2}[/tex]
[tex]200\times(3958.756)^2=Gmm_{e}[/tex]
[tex]Gmm_{e}=3.134\times10^{9}[/tex]
We need to calculate the height of the object
Using formula of gravitational force
[tex]F=\dfrac{Gmm_{e}}{r^2}[/tex]
Put the value into the formula
[tex]125=\dfrac{200\times(3958.756)^2}{r^2}[/tex]
[tex]r^2=\dfrac{200\times(3958.756)^2}{125}[/tex]
[tex]r^2=25074798.5[/tex]
[tex]r=\sqrt{25074798.5}[/tex]
[tex]r=5007.4\ miles[/tex]
Hence. The height of the object is 5007.4 miles.
What happens to the temperature of a substance while it is changing state?
Answer:
its temperature stays constant
Explanation:
Oil is a powerful source of energy used for cars, machines, and many other purposes. While studying sources of energy that could replace the use of oil, a student thinks about solar power, wind power, nuclear power, and electric batteries. The students asks, "Which type of source will waste the least energy and provide the most energy in all situations?" Which best describes why a scientist would ask a different question than this one?
The question asks about subjective personal preferences regarding energy.
The question focuses on the objective measurements of amounts of energy.
The question has too wide of a focus, as though all situations can be studied at once.
The question is focused on the future and finding new answers about changing needs.
Answer:
The question has too wide of a focus, as though all situations can be studied at once
Explanation:
I took the test
Answer:
The question has too wide of a focus, as though all situations can be studied at once
Explanation:
i took ed test
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)
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
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
We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (human body can withstand very large amount of impulse, if it's delivered over a long time with small forces, but we cannot withstand very large forces lasting over more than a few milliseconds, delivering relatively small impulse). In order to estimate average force from impulse, we need the duration of interaction. Suppose that the contact of the bat with the baseball in (b) lasts for 0.7 milliseconds. What is the magnitude of the average force that the bat exerts on the baseball, for the duration of contact
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]I = 476 \ N \cdot s [/tex]
b
[tex]I_1 = 14.21 \ N\cdot s [/tex]
c
[tex]F = 20300 \ N [/tex]
Explanation:
Considering the first question
From the question we are told that
The force produced is [tex]F = 3400 \ N[/tex]
The duration of the punch is [tex]t = 0.14 \ s[/tex]
Generally the impulse delivered is mathematically represented as
[tex]I = F * t[/tex]
=> [tex]I = 3400 * 0.14[/tex]
=> [tex]I = 476 \ N \cdot s [/tex]
Considering the second question
The approaching velocity of the ball is [tex]v_b = 45 \ m/s[/tex]
The leaving velocity of the ball is [tex]v_l = -53 \ m/s[/tex]
The mass of the ball is [tex]m_b = 0.145 \ kg[/tex]
Generally the magnitude of the impulse delivered is mathematically represented as
[tex]I_1 = m* v_b - m * v_l[/tex]
=> [tex]I_1 = [0.145 * 45] - [0.145 * -53][/tex]
=> [tex]I_1 = 14.21 \ N\cdot s [/tex]
Considering the third question
The duration of the impact of the bat is [tex]t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s[/tex]
Generally the average force exerted by the bat is mathematically represented as
[tex]F = \frac{I_1}{t_1}[/tex]
=> [tex]F = \frac{14.21 }{0.7 *10^{-3}}[/tex]
=> [tex]F = 20300 \ N [/tex]
