If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
ANSWER AND I WILL GIVE YOU BRAINILIEST
Answer:
73N
Explanation:Just multiply 1.2^2 by 50
A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
Learn more about banking: https://brainly.com/question/12134554
I’ll mark brainless please hurry
Answer:
Covalent bonds can form between similar atoms.
Explanation:
A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 9.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.80 times that due to gravity
Answer:
13.37 rev/min
Explanation:
acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².
r = 9 m
Centripetal acceleration ([tex]a_c[/tex]) is given by:
[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]
The velocity (v) is given by:
v = ωr; where ω is the angular velocity
Hence:
ω = v/r = 12.6 / 9
ω = 1.4 rad/s
ω = 2πN
N = ω/2π = 1.4 / 2π
N = 0.2228 rev/s
N = 13.37 rev/min
Calculate the magnitude and direction of the resultant of the following forces
Answer and I will give you brainiliest
Answer:
250N same direction
Explanation:
100+150 = 250N
same direction
An aircraft flies 800km due East and then 600km due north. Determine the magnitude of it's displacement. please it's urgent
Answer:
Magnitude of it's displacement = 1,000 km
Explanation:
Given:
Distance towards east = 800 km
Distance towards north = 600 km
Find:
Magnitude of it's displacement
Computation:
Magnitude of it's displacement = √800² + 600²
Magnitude of it's displacement = √640000 + 360000
Magnitude of it's displacement = √10000000
Magnitude of it's displacement = 1,000 km
What impulse must be applied to a 25.0-kg cart to cause a velocity change
of 12.0 m/s?
Answer:
Impulse of force = 300Ns
Explanation:
Given the following data;
Mass = 25kg
Change in velocity = 12m/s
To find the impulse;
Impulse is given by the formula;
[tex] Impulse \; of \; force = mass * change \; in \; velocity [/tex]
Substituting into the equation, we have;
[tex] Impulse \; of \; force = 25 * 12 [/tex]
Impulse of force = 300Ns
A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.
Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?
Answer:
Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2
Explanation:
See the attached files.
To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.
Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.
The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.
(a) What type of charge distribution is inside the surface? a positively charged plane parallel to the end faces of the cylinder a positive line charge situated on and parallel to the axis of the cylinder a collection of positive point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a collection of negative point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a negatively charged plane parallel to the end faces of the cylinder (b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? N · m2/C (c) What is the net charge inside the cylinder? nC
Answer:
a) The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.
b) the net electric flux is -821.09 N.m²/c
c) the net charge inside the cylinder is -7.2666 nC
Explanation:
Given the data in the question;
a) (a) What type of charge distribution is inside the surface?
Based on the Image of the question below;
The type of charge distribution inside the surface is a negatively charged plane parallel to the end faces of the cylinder.
b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface?
we know that; Electric flux is;
∅ = -2 × E × Area
we substitute
∅ = -2 × 300 × π ( 0.66 m )²
= -821.09 N.m²/c
Therefore; the net electric flux is -821.09 N.m²/c
c) What is the net charge inside the cylinder? nC
from Gause's law;
∅ = q/∈₀
q = ∅ ∈₀
we know permittivity ∈₀ = 8.85 × 10⁻¹²)
we substitute
q = -821.09 N.m²/c × (8.85 × 10⁻¹²)
q = -7.2666 × 10⁻⁹ C
q = -7.2666 nC
Therefore, the net charge inside the cylinder is -7.2666 nC
What is one example of an individual in an ecosystem?
Answer:
a
Explanation:
AAAA
A certain green light bulb emits at a single wavelength of 550 nm. It consumes 55 W of electrical power and is 75% efficient in converting electrical energy into light. (a) How many photons does the bulb emit in one hour? (b) Assuming the emitted photons to be distributed uniformly in space, how many photons per second strike a 10 cm by 10 cm paper held facing the bulb at a distance of 1.0 m?
Answer:
a) #_total = 4 10²³ photons / h, b) # _photon_area = 3 10²² photons
Explanation:
a) Let's start by calculating the energy of an emitted photon
E₀ = h f
c = λ f
substituting
E₀ = h c /λ
E₀ = [tex]\frac{6.62 \ 10^{-34} \ 3 \ 10^{8} }{550 \ 10^{9} }[/tex]
E₀ = 3.6 10⁻¹⁹ J
Let's use a direct proportion rule (res rule) if a photon has Eo, how many photons are there in 55W
#_foton = 55 / E₀
# _photon = 55 / 3.6 10⁻¹⁹ = 15.27 10¹⁹ photons
This version is with 100% if the conversion is 75%, how many footnes are there
#_foton_real = # _foton 75/100
#real_photon = 15.27 10¹⁹ 0.75
# _real_photon = 11.45 10¹⁹ photons
this is the broadcast in a second
#_total = # _real_photon t
#_total = 11.45 10¹⁹ 3600
#_total = 4.1 10²³ photons / h
#_total = 4 10²³ photons / h
b) This number of photons is constant, so after being emitted they are distributed on the surface of a sphere, in this case of radius r = 1.0 m
the volume of a sphere is
A = 4π r²
A = 4π 1²
A = 12,566 m²
the area of the plate is
A₁ = l₁ l₂
A₁ = 0.10 0.10
A₁ = 1 10⁻² m
Let's use a direct proportion rule, if there are 4.1 10²³ photons in an area A, how much are there in an area A₁
# _photon_area = #_total A₁ / A
# _photon_area = [tex]4.1 \ 10^{23} \ \frac{1 \ 10^{-2} }{12.566}[/tex]
# _photon_area = 3.26 10²²
as the number of photons must be a whole number
# _photon_area = 3 10²² photons
A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.
Answer:
31.68 m/s
Explanation:
The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.
As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:
[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]
Now we just add the time accelerating:
[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]
A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?
Answer:
eeeeeeeeeeeeeeeeeeeee
Sound waves are what type of energy wave?
A. mechanical
B. longitudinal
C. transverse
Answer:
mechanical wave
Explanation:
The sound vibrations cause waves of pressure that travel through a medium, such as air, water, wood or metal. Sound energy is a form of mechanical energy
Answer:
A mechanical waves
Explanation:
in nature include water waves, sound waves, seismic waves and internal water waves
A star's emission line of 400 nm appears shifted to 404 nm in the spectrum. What can you conclude from this shift?
A. The star is approaching you with the speed of 3000 km/s.
B. The star is approaching you with the speed of 30300 km/s.
C. The star is receding from you with the speed of 3000 km/s.
D. The star is receding from you with the speed of 30300 km/s.
Answer:
C. The star is receding from you with the speed of 3000 km/s
Explanation:
To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.
We solve for V
V = 3x10⁶m/s
V = 3000km/s
We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.
Thank you!
The open-circuit voltage of a car battery is measured to be 12 V. During engine startup, the battery delivers 10 A to the starter motor, and the battery voltage drops to 11.7 V. Draw the Thévenin equivalent circuit for the battery. How much power does the battery deliver to the starter motor?
Answer:
- the Thevenin equivalent circuit for the battery is uploaded below
- the battery delivered 117 watts of power to the starter motor
Explanation:
Given the data in the equation
diagram of the Thevenin equivalent circuit for the battery is uploaded below.
Current I = 10 A
Voltage 1 = 12 V
voltage 2 = 11.7 v
R = (V1 - V2) / I
R = (12-11.7)/10
R = 0.3 / 10
R = 0.03Ω
Thevenin equivalent circuit
[tex]R_{L}[/tex] = V2 / I = 11.7 / 10
[tex]R_{L}[/tex] = 1.17Ω
so, power delivered to the starter motor will be;
p = (V2)² / [tex]R_{L}[/tex]
P = ( 11.7 V )² / 1.17Ω
p = 136.89 / 1.17
p = 117 watts
Therefore, the battery delivered 117 watts of power to the starter motor
In a series circuit of two resistors, the resistors have the same __________; in a parallel circuit of two resistors, the resistors have the same __________. In a series circuit of two resistors, the resistors have the same __________; in a parallel circuit of two resistors, the resistors have the same __________. Voltage across them, current through them Current through them, voltage across them Current through them, current through them Voltage across them, voltage across them
Answer:
Current through them, voltage across them.
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
[tex] V = IR[/tex]
Where;
V represents voltage measured in voltage.
I represents current measured in amperes.
R represents resistance measured in ohms.
A resistor can be defined as an electronic component that opposes the free flow of current from one point to another in an electrical circuit.
In a series circuit of two resistors, the resistors have the same current through them; in a parallel circuit of two resistors, the resistors have the same voltage across them.
The all-digital touch-tone phones use the summation of two sine waves for signaling. Frequencies of these sine waves are defined as 697, 770, 852, 941, 1209, 1336, 1477, and 1633 Hz. Since the sampling rate used by the telecommunications is 8000 Hz, convert those eight analog frequencies into digital frequencies of radians and cycles.
HELP ! ILL MARK BRAINLIEST HELP ASAP
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!
Four point charges each having charge Q are located at the corners of a square having sides of length a. Find symbolic expressions for the work required to bring a fifth charge q from infinity to the center of the square.
Answer:
Explanation:
We shall first calculate electric potential at the centre of square due to four charges of Q .
potential = k Q / r , k = 9 x 10⁹ , Q is charge which is placed at distance r from the point where potential is measured .
Distance of corner of square to centre = √ 2 a / 2 = a / √2 = .707 a
potential due to one charge
= k x Q / .707 a
= 9 x 10⁹ x Q / .707a
= 12.73 x 10⁹ Q/a
Potential due to 4 charges
= 4 x 12.73 x 10⁹ Q/a
= 50.92 x 10⁹ Q/a .
Potential at infinity = 0
Work done in carrying charge q from infinity to centre of square
= potential difference x charge
= ( 50.92 x 10⁹ Q/a - 0 ) x q
= 50.92 x 10⁹ x Qq / a J .
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
Answer:
22
Explanation:
before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv
3. A stone is thrown vertically upwards from the top of a building 50 m tall with an initial
velocity of 20.0 ms. If the stone just misses the edge of the roof on its return, determine
(a) The time is taken the stone to get to its maximum height.
(b) The maximum height reached by the stone
(c) The time at which the stone return to the point where it was thrown
(d) The velocity of the stone at this instance
(e) The velocity and position of the stone at t = 5 s.
Answer:
13.4436
Explanation:
A ball is thrown horizontally to the right, from the top of a vertical cliff of height h. A wind blows horizontally to the left, and assume (simplistically) that the effect of the wind is to provide a constant force to the left, equal in magnitude to the weight of the ball. How fast should the ball be thrown so that it lands at the foot of the cliff
Answer:
v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]
Explanation:
For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
this time is the same as the horizontal movement.
Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air
F = m aₓ
they tell us that force is equal to the weight of the body
-mg = maₓ
aₓ = -g
the sign indicates that the acceleration is to the left
we write the kinematics equation
x = x₀ + v₀ₓ t + ½ aₓ t²
They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v
we substitute
0 = 0 + v t + ½ (-g) t²
v = ½ g t
we use the drop time
v = ½ g [tex]\sqrt{\frac{2yo}{g} }[/tex]
v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]
The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?
Answer:
θ = 90º
Explanation:
The velocity is given by
v = [tex]\frac{dr}{dt}[/tex]
calculate
v = 3 i ^ + √2 j ^ + 2t k ^
acceleration is defined by
a = dv / dt
a = 2 k ^
one way to find the angle is with the dot product
v. a = | v | | a | cos θ
cos θ= v.a / | v | | a |
Let's look for the value of each term
v. a = 4 t
| v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]
| a | = 2
they ask us for the angle for time t = 0
v. a = 0
| v | = √11 = 3.317
we substitute
cos θ = 0 /√11
cos θ = 0
therefore the angles must be θ = 90º
A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey
Answer:
1.875 x 10⁶ m /s .
Explanation:
Force on electron = E e where E is electric field and e is charge on electron
acceleration generated = Ee / m where m is mass of the electron .
Putting the values
acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹
= .879 x 10¹² m /s²
v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m
v² = 0 + 2 x .879 x 10¹² x 2
v = 1.875 x 10⁶ m /s .
Which of the following is not a true statement?
A
B
C
D
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
First to answer gets brainliest
Answer:
has a charge
Explanation:
Answer:
gain electrons
Explanation:
I looked it up and that was what I found
side note: ions do have a charge
As the distance between the sun and earth decreases, the speed of the planet
a
increases
b
decreases
c
stays the same
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
Earth has seasons because _____.
it rotates on its axis as it moves around the sun
the temperature of the sun changes
its axis is tilted
the distance between Earth and the sun changes
Answer:
c, its axis is tilted
maybe
As it works its way around the sun, its tilted axis exposes different parts of earth.
C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis