Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
Help solve these two problems im having trouble trying to start these problems?
Answer:
25. Approximately 8.1 meters
26. North 1.31 km, and East 2.81 km
Explanation:
25.
Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:
[tex]d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m[/tex]
which can be rounded to 8.1 m.
26.
Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.
We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):
[tex]east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km[/tex]
and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.
[tex]north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km[/tex]
I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 75 kg skier travels downhill 1200 meters in 56 seconds. What is the velocity of the skier?
Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
An 80-kg bungee jumper jumps off a bridge. Rubber bungee cords act as a large spring attaching the jumper to the bridge. A bear standing in the river below catches the jumper. If the spring constant of the bungees is 20 N/m and they stretch 50 m. How much force must the bear apply to keep the jumper from moving?
Answer:
The force is [tex]F_b = 216 \ N [/tex]
Explanation:
From the question we are told that
The mass of the bungee jumper is m = 80 kg
The spring constant is [tex]k = 20 \ N/ m[/tex]
The extension of the rubber bungee cords is x = 50 m
Generally the weight of the jumper is
[tex]W = m * g[/tex]
=> [tex]W = 80 * 9.8 [/tex]
=> tex]W = 784 \ N [/tex]
Generally the returning force of the rubber bungee cords is mathematically represented as
[tex]F = k * x[/tex]
=> [tex]F = 20 * 50 [/tex]
=> [tex]F = 1000 \ N [/tex]
The force to be applied by the bear is
[tex]F_b = F - W[/tex]
=> [tex]F_b = 100 - 784[/tex]
=> [tex]F_b = 216 \ N [/tex]
The word acid comes from the Latin word
If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *
A. distance between two falling objects will keep increasing
B. distance between two falling objects will keep decreasing
C. will stay the same
D. All of the above
Answer:
A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.
Explanation:
. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?
Answer:
The value is [tex]n= 9.375 *10^{15} \ electrons [/tex]
Explanation:
From the question we are told that
The average current is [tex]I = 25.0 \mu A = 25.0 *10^{-6} \ A[/tex]
Generally the quantity of charge (electron ) is mathematically represented as
[tex]Q = ne[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally current is mathematically represented as
[tex]I = \frac{Q}{t}[/tex]
=> [tex]I = \frac{ne}{t}[/tex]
Here t is time which is given as 1 minutes = 60 seconds
and n is the number of electrons
So
[tex]25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}[/tex]
=> [tex] 60 * 25.0 *10^{-6} = n* 1.60 *10^{-19} [/tex]
=> [tex]n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }[/tex]
=> [tex]n= 9.375 *10^{15} \ electrons [/tex]
The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons
What is the number of electrons?
We are given;
Average Current; I = 25 μA = 25 × 10⁻⁶ A
Formula for the current is;
I = Q/t = ne/t
where;
n is number of electrons
e is electron charge = 1.6 * 10⁻¹⁹ C
t is time = 1 minute = 60 seconds
Thus making n the subject gives;
n = It/e
n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)
n = 9.375 × 10¹⁵ electrons
Read more about number of electrons at; https://brainly.com/question/11406294
A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?
Answer:
t = 5.77 s
Explanation:
This exercise will use Newton's second law for rotational motion
τ = I α
α = τ / I
α = 66/175
α = 0.3771 rad/s²
now we can use the rotational kinematics relations, remember that all angles must be in radians
θ = 1 rev = 2π radians
θ = w₀ t + ½ α t²
as the wheel starts from rest w₀ = 0
t = √ (2θ/α)
let's calculate
t = √ (2 2π / 0.3771)
t = 5.77 s
A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this to m/s). What is the minimum coefficient of static friction needed to keep the tires from slipping? Explain why the answer is (or is not) plausible.
Answer:
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
Explanation:
From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)
[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]m[/tex] - Mass of the motocyclist, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.
[tex]R[/tex] - Radius of the circular turn, measured in meters.
The static coefficient of friction is cleared in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]
From (Eq. 2) we get that normai force is:
[tex]N = m\cdot g[/tex]
And we expand the resulting expression in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]
[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)
If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:
[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2
Answer:
hi
Explanation:
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Velocity which stone gains when falling from height of 80 m is approximately equal to *
A. 0
B. 1 m/s
C. 8 m/s
D. 40 m/s
E. 300 m/s
Answer:
40
Explanation:
Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?
Answer:
3.2
Explanation:
I hope that this helps! Have a good day!!
I need help with this answer
Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds to 212∘F, calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Give your answer to two significant figures.
Answer:
In two significant figure 360K
Explanation:
The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.
Hence,
ΔT = 212 - 32
ΔT = 180°F
To convert to °F to kelvin, we use the formula below
= (°F - 32) × 5/9 + 273.15
= (180°F - 32) × 5/9 + 273.15
= 355.37K ⇔ 360K
Find the mass of the two-dimensional object. A disk of radius 7 in with density at distance x from origin given by rho ( x ) = √ 10 x
Answer:
1030.3504 units
Explanation:
Disk radius = 7 inches
Density at distance x = p(x) = [tex]\sqrt{10x}[/tex]
Calculate the mass of the two-dimensional object
M = 2[tex]\pi[/tex][tex]\sqrt{10} * (\frac{2}{5} ) *(7)^{\frac{5}{2} }[/tex] = 1030.3504 units
Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.
Answer:
1,839.375 Joules
Explanation:
Work is said to be done is the force applied to an object cause the object to move through a distance.
Workdone = Force * Distance
Workdone = mass * acceleration due to gravity * distance
Given
Mass = 75.0kg
acceleration due to gravity = 9.81m/s²
distance = 2.50m
Substitute the given parameters into the formula:
Workdone = 75.0*9.81*2.50
Workdone = 1,839.375Joules
Hence the workdone is 1,839.375 Joules
1. Name three branches of
science.
Answer:
Biology
Physics
Chemistry
Explanation:
Biology (life like cells, human reproduction, etc.)
Physics (Studies forces, like gravity.)
Chemistry (studies the atoms, the elements, etc.)
A student studies the effect of an object's speed on its amount of kinetic energy. This graph summarizes the data from the study Kinetic energy Speed Which statement best describes what the graph shows?
A. As speed increases, kinetic energy increases exponentially
B. As speed increases, kinetic energy stays the same
C. As speed decreases, kinetic energy doubles each time.
D. As kinetic energy increases, speed decreases exponentially
The answer is A I Hope this answer helps you (i got the question right)
Answer:
A. As speed increases, kinetic energy increases exponentially.
Explanation:
The amount of kinetic energy an object has depends on the speed. Kinetic energy is also known as "motion energy." This being said, if speed is increasing, decreasing, or staying constant, the kinetic energy of the object will too.
6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.
Answer and Explanation:
TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.
[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]
The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).
Numbers:
Sun’s mass: 2 x 10^30 kg
Neptune’s mass: 1 x 10^26 kg
Distance of Sun to Uranus: 3 x 10^9 km
Closest approach of Uranus and Neptune: 1.5 x 10^9 km
Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.
The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.
The formula for calculating the Force of Gravity between two masses is:
F = G*m₁*m₂/r²
Where;
F = force of gravity
G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²
m₁ = mass of the larger object
m₂ = mass of the smaller object
r = the distance between the centers of the two masses
Now, from online values, we have the following;
mass of Neptune; m₁ = 102.413 × 10²⁴ kg
mass of Uranus; m₂ = 86.813 × 10²⁴ kg
average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m
Thus, force exerted by Neptune on Uranus is;
F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²
F = 2.240 × 10¹⁷ N
We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.
That is greater than the force Neptune exerts on Uranus.
Read more about Force of Gravity at; https://brainly.com/question/7281908
An ideal gas increases in temperature from 22°C to 42°C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which process requires more heat or are the required amount of heat same in both?
Answer:
a- More heat is required for the constant-pressure process than for the constant-volume
Explanation:
we have to solve using the thermodynamic first law. this is the heat applied to the system
dQ = dU + dW
definition of terms:
dU = change in internal energy
dW = work done
we have it that
change in internal energy dU is directly proportional to work done dW
but when we are in constant volume process, work done of the gas is zero
therefore
dQ of constant pressure is > than that of constant volume
so constant pressure process requires more heat
The process that requires more heat is the constant-pressure process than the constant-volume process.
According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.
In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.
Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.
Read related link on:
https://brainly.com/question/16951562
How many strings of length 10 over the alphabet (a, b, c, d] have at least one b somewhere in the string?
a) 310
b) 410 - 310
c) 10.4
d) 10.39
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is B
Explanation:
The number of alphabet is n= 4 (a , b , c , d )
Generally the total number of string of length 10 over the 4 alphabets is
[tex]N = 4^{10}[/tex]
Gnerally the number of string of length 10 that does not include b is
[tex]T = 3^{10}[/tex]
Generally the number of strings of length 10 over the 4 alphabets that have at least one alphabet b somewhere in the string is
[tex]G = N - T[/tex]
=> [tex]G = 4^{10} - 3^{10}[/tex]
A single living thing.
Answer:
What do you mean ma´am/sir?
Explanation:
as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is
Answer:
The value of q is [tex]\dfrac{Q}{8}[/tex]
Explanation:
Given that,
Each charge = -Q
Distance between charges = L
Reduced force = [tex]\dfrac{F}{2}[/tex]
Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.
We know that,
The force on each end is
[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)
If the charge q is placed at mid point then
The force on each end charge is
[tex]\dfrac{F}{2}=F+F'[/tex]....(II)
We need to calculate the value of q
Using equation (II)
[tex]\dfrac{F}{2}=F+F'[/tex]
Put the value of F into the formula
[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]
[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]
[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]
[tex]q=\dfrac{Q}{8}[/tex]
Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]
what is the meaning of relative as a noun?
Answer:
noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?
Answer:
b
Explanation:
How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?
Answer:
The answer is 1450 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 290 N
distance = 5 m
We have
workdone = 290 × 5
We have the final answer as
1450 JHope this helps you
A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.011 m . Part A Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest. Express your answer using three significant figures and include the appropriate units.
Answer: IM 95%sure that the answer is B jus took the test got the answer right
Explanation:
Answer:
sorry I forgot I wish I could help
Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answer the following questions. You hold a small metal ball of mass a height above the floor. You let go, and the ball falls to the floor. Choose the origin of the coordinate system to be on the floor where the ball hits, with up as usual. Express all results in terms of , , and . Just after release, what are and
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration
A region around the nucleus of an atom where electrons are likely to be found
Answer:
The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.
Explanation:Hope this helps :)
By definition, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
First of all, an atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.
All atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged.
In other words, every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.
This is, the atomic nucleus is the central part of the atom that is made up of protons and neutrons, while the orbitals or peripheral region is an area where electrons are found.
In summary, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
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