Answer:
T = 98 N
Explanation:
The gravity of the earth is known to be 9.8 m/s²
Data:
m = 10 kgg = 9.8 m/s²T = ?Use formula:
[tex]\boxed{\bold{T=m*g}}[/tex]Replace and solve:
[tex]\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{T=98\ N}}}[/tex]The tension in the rope is 98 Newtons.
Greetings.
Calculate the spring constant of the spring in a child's pogo stick if the child has a mass of 32 kg and bounces once every 2.7 seconds
Answer:
The spring constant is 173.12 N/m.
Explanation:
mass, m = 32 kg
Period, T = 2.7 s
let the spring constant is K.
Use the formula of period,
[tex]T = 2\pi\times \sqrt \frac{m}{K}\\\\2.7 =2\times 3.14\sqrt\frac{32}{K}\\\\K = 173.12 N/m[/tex]
A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?
Answer:
Answer is 183.6 J
Explanation:
Using the Physics reference sheet the formula for Potential energy is
(mass) x (gravity) x (height)
Mass= 1.2
Gravity I used is 9.81 (use 10 to get the answer most schools use)
Height= 15.6
The function s(t)s(t) describes the position of a particle moving along a coordinate line, where ss is in feet and tt is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, specd, and acceleration at time t
Answer:
Explanation:
From the given information:
Let's assume that the missing function is:
s(t) = t³ - 6t², t ≥ 0
From part (b), we are to find the given required terms when time t = 2
So; from the function s(t) = t³ - 6t², t ≥ 0
[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]
[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]
[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]
[tex]acceleration\ a(t) = 6t - 12[/tex]
At time t = 2
The position; S(2) = (2)² - 6(2)²
S(2) = 8 - 6(4)
S(2) = 8 - 24
S(2) = - 16 ft
v(2) = 3(2)² - 12 (2)
v(2) = 3(4) - 24
v(2) = 12 - 24
v(2) = - 12 ft/s
speed = |v(2)|
|v(2)| = |(-12)|
|v(2)| = 12 ft/s
acceleration = 6t - 12
acceleration = 6(2) - 12
acceleration = 12 - 12
acceleration = 0 ft/s²
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = [tex]\frac{x-x_1}{t}[/tex]
we substitute the values
v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = [tex]\frac{6600 - x_1}{4}[/tex]
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = [tex]\frac{6600 -128 a}{4}[/tex]
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
The diagram shows a transistor used in a circuit.
A signal with a small change in voltage is input into a circuit that includes a transistor connected to 4 resistors and 2 capacitors. The signal output from the transistor has a much larger change in voltage than the input.
What does the diagram show?
an NPN transistor used as an amplifier
an NPN transistor used as a switch
a PNP transistor used as an amplifier
a PNP transistor used as a switch
Answer:
a PNP transistor used as an amplifier
Explanation:
The diagram show a PNP transistor used as an amplifier.PNP transisitor having one 2 P type and the 1 N type of semiconductor.
What is PNP transistor?This bipolar PNP junction transistor is made up of three layers of semiconductor material, two of which are P-type & one of which is N-type. It consists of three terminals.
The transistor's emitter allows it to supply the majority of charge carriers. In relation to the ground, the emitter always is forward biased.
As a result, the base receives the vast majority of charge carriers. A transistor's emitter is strong and of modest size.
The collector collects the vast majority of the charge carrier delivered by the emitter. Reverse bias is always present at the collector-base junction.
The charge collector region is moderately mixed and capable of collecting the charge.
The diagram shows a transistor used in a circuit.
A signal with a small change in voltage is input into a circuit that includes a transistor connected to 4 resistors and 2 capacitors. The signal output from the transistor has a much larger change in voltage than the input.
The diagram show a PNP transistor used as an amplifier.
Hence option C is correct.
To learn more about the PNP transistors refer to the link;
https://brainly.com/question/1492057
Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?
u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.
Given-
velocity of the ice skater is 12 m/ sec.
Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.
[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]
The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,
[tex]W_{f}=\bigtriangleup KE +0[/tex]
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,
[tex]W_{f} =u_{f} mgx[/tex]
Here, [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .
Equate the value of kinetic energy and work done of friction for further result, we get,
[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]
[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]
[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]
[tex]u_{f} =0.077[/tex]
Hence, the coefficient of kinetic friction between the ice and skates is 0.077.
For more about the friction, follow the link below-
https://brainly.com/question/13357196
True or false, wrrect the false
statement:
• The magnetic field created by a flat coil is
uniform.
• Inside a solenoid, the lines of field are
oriented from the north face to the south
face.
• The magnetic field outside Helmholtz
coils is uniform.
• Le champ B à l'intérieur d'un solénoïde
est uniforme.
• The magnitude of B, created by a flat coil
of radius R, at any point in its plane is
B= 2m x 10-NI
R
• The designation of the faces of a wil
depend the sense of the current
traversing it.
Answer:
false
Explanation:
which statement regarding the idealized model of motion called free fall is true?
a. the effect of air resistance is factored in the equation of motion in the idealized model called free fall.
b. free fall only models motion for objects that do not have an initial velocity in the upward direction.
c. the idealized model of the motion called free fall applies in cases where distance of the fall is large compared with the radius of the astronomical body on which the fall occurs.
d. a freely falling object has a constant acceleration due to gravity.
Cuando Daniel hace oscilar un péndulo, este realiza 30,6 ciclos (completos) en 9 [s].
¿Cuál es la frecuencia del péndulo?
A )3,4 [Hz].
B )4,3 [Hz].
C )30 [Hz].
D )5 [Hz]
If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process. Express your answer in volts.
The question is incomplete. The complete question is :
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.
Solution :
Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.
It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.
Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :
[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex] and [tex]$\phi_{B,f} = 0$[/tex]
Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,
[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]
[tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]
= 30.27 V
Therefore, the emf generated is 30.27 V.
In a double-slit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) Calculate the angular separation, !, in radians between the central maximum and the 1st order maximum
Solution :
The conditions for the maximum in the Young's experiment is :
d sin θ = m λ, where m = 0, 1, 2, 3, .....
The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,
d sin θ = λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]
Given : d = 100 λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]
[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]
[tex]$=0.573^\circ$[/tex]
= 0.01 rad
A ship moves 330 m after 40,000 j of work is done on it. What force is used to do this work?
Explanation:
F=w/s
F=40,000j/330m
F=121.212N
An electric field has a positive test charge of 5.00 C placed in it. The force on the test charge is
6.000 N. The magnitude of the electric field at the location of the test charge is
o 30.0 NVC
0 1.20 N/C
0 120, NVC
O 3.00 N/C
01.02 N/C
Answer:
yes
Explanation:
this means the answer is yes
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
A.
69.58 newtons
B.
45.05 newtons
C.
25 newtons
D.
98.26 newtons
Answer:
Option A (69.56 newtons) is the appropriate solution.
Explanation:
According to the question,
On the X-axis,
⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]
or,
[tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]
On substituting the values, we get
[tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]
[tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)
On the Y-axis,
⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]
[tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]
[tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]
From equation 1, we get
[tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]
[tex]T_1+3T_1=278.4 \ N[/tex]
[tex]4T_1=278.4 \ N[/tex]
[tex]T_1=\frac{278.4}{4}[/tex]
[tex]=69.6 \ N[/tex]
Answer:
69.58
Explanation:
16. Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater's nose it
it is 0.120 m from the axis of rotation?
Answer:
a) w = 31.4 rad / s, b) a = 118.4 m / s²
Explanation:
a) let's reduce to the SI system
w = 5 rev / s (2pi rad / 1 rev)
w = 31.4 rad / s
b) the expression for the centripetal acceleration is
a = v² / r
linear and angular variables are related
v = w r
we substitute
a = w² r
a = 31.4² 0.120
a = 118.4 m / s²
pls can anyone solve this
Answer:
3 pls give me brainliest
Explanation:
A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
ocean floor and is received 15s later. What is the distance between the submarine and the ocean floor?
Answer:
the distance between the submarine and the ocean floor is 11,250 m
Explanation:
Given;
speed of the wave, v = 1500 m/s
time of motion of the wave, t = 15 s
The time taken to receive the echo is calculated as;
[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]
Therefore, the distance between the submarine and the ocean floor is 11,250 m
a train is traveling at 50km/h average .what is the displacement of the train per second?
A 6 kg ball experiences a 5 m/s^2 acceleration. What is the strength of the force felt by the ball?
a: 0.83kg
b: 30 newtons
c: 30 kg
d: 1.2 newtons
Answer:
30 newtons
explanations
data given
mass=6kg
acceleration=5
f=m×a
6×5=30
Your friend has been given a laser for her birthday. Unfortunately, she did not receive a manual with it and so she doesn't know the wavelength that it emits. You help her by performing a double-slit experiment, with slits separated by 0.36 mm. You find that the two m n = 2 bright fringes are 5.5 mm apart on a screen 1.6 m from the slits.
a. What is the wavelength the light emits?
b. What is the distance between the two n = 1 dark fringes?
Answer:
a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m
b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m
Explanation:
Given the data in the question;
separation between two slits d = 0.36 mm = 0.00036 m
Separation between two adjacent fringes β = 5.5 mm = 0.0055 m
Distance of screen from slits D = 1.6 m
n = 2
a) the wavelength the light emits;
Using the formula;
β = (nD/d)λ
To find wavelength, we make λ the subject of formula;
βd = nDλ
λ = βd / nD
so we substitute
λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )
λ = 0.00000198 / 3.2
λ = 6.1875 × 10⁻⁷ m
Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m
b) the distance between the two n = 1 dark fringes;
To find the distance between the two n = 1 dark fringes, we use the following formula;
y[tex]_m[/tex] = 2nλD / d
given that n = 1, we substitute
y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m
y[tex]_m[/tex] = 0.00000198 / 0.00036
y[tex]_m[/tex] = 0.0055 m
y[tex]_m[/tex] = 5.5 × 10⁻³ m
Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9m2 .
Answer:
0.9378
Explanation:
Weight (W) of the rider = 100 kg;
since 1 kg = 9.8067 N
100 kg will be = 980.67 N
W = 980.67 N
At the slope of 12%, the angle θ is calculated as:
[tex]tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0[/tex]
The drag force D = Wsinθ
[tex]\dfrac{1}{2}C_v \rho AV^2 = W sin \theta[/tex]
where;
[tex]\rho = 1.23 \ kg/m^3[/tex]
A = 0.9 m²
V = 15 m/s
∴
Drag coefficient [tex]C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}[/tex]
[tex]C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}[/tex]
[tex]C_D =0.9378[/tex]
4. How much milk at 5° C needs to be added to 250 g of coffee at 90° C to make the coffee drinkable at 60° C?
Answer:
dino :)
Explanation:
Planet K2-116b has an Average orbital radius of 7.18x10^9 m around the star K2-116. It has a mass of about 0.257 times the mass of the earth and an orbital period of 2.7 days.
What is the orbital speed of the planet?
Determine the mass of the star.
a) v = 1.94 × 10^5 m/s
b) Ms = 2.09 × 10^24 kg
Explanation:
Given:
m = 0.257M (M = mass of earth = 5.972×10^24 kg)
= 1.535×10^24 kg
r = 7.18×10^9 m
T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)
= 2.3328×10^5 s
a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:
C = 2×(pi)×r
= 2(3.14)(7.18×10^9m)
= 4.51×10^10 m
The orbital speed v is then given by
v = C/T
= (4.51×10^10 m)/(2.33×10^5 s)
= 1.94 × 10^5 m/s
b) We know that centripetal force Fc is given by
Fc = mv^2/r
where v = orbital speed
r = average orbital radius
m = mass of planet
We also know that the gravitational force FG between the star K2-116 and the planet is given by
FG = GmMs/r^2
where m = mass of planet
Ms = mass of star K2-116
r. = average orbital radius
G = universal gravitational constant
= 6.67 × 10^-11 m^3/kg-s^2
Equating Fc and FG together, we get
Fc = FG
mv^2/r = GmMs/r^2
Note that m and one of the r's get cancelled out so we are left with
v^2 = GMs/r
Solving for the mass of the star Ms, we get
Ms = rv^2/G
=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)
= 2.09 × 10^24 kg
Give 2 reasons for fitting heavy commercial vehicles with many tyres
As we know larger the area of contact lesser the pressure. So, in order to reduce the pressure heavy vehicles have broad tyres to increase the area of contact with the ground. Heavy vehicles have broad tyres because broad tyres have large area of contact and less pressure on the ground.
mark me brainliesttt pls :)))
Which of the following would likely happen if a person’s lactic acid system had difficulty breaking down glycogen in the muscles?
The person would have difficulty swimming across a lake.
The person would have difficulty sprinting in a race.
The person would have difficulty cycling down a hill.
The person would have difficulty running a marathon.
Answer: I think that its b, they would have difficulty sprinting in a race
Explanation:
A deer with a mass of 156 kg is running head on toward you with a speed of 10 m/s. Find the momentum of the deer
Hi there!
[tex]\large\boxed{1560 kgm/s}[/tex]
Recall that:
P = m · v, where:
P = momentum
m = mass (kg)
v = velocity (m/s)
Thus:
P = 156 · 10
P = 1560 kgm/s
what is the meaning of friend ?
Answer:
person that you know and like (not a member of your family), and who likes you
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.175 m^3 at a pressure of 0.350 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.365 m^3.
Required:
If the temperature remains constant, what is the final value of the pressure?
Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
a. Calculate the focal length of the mirror formed by the shiny bottom of a spoon that has a 2.51 cm radius of curvature.
Answer:
f = 1.255 cm
Explanation:
The Radius of Curvature:
The radius of that hollow sphere, whose part is the spherical mirror, is known as ‘The Radius of Curvature’ of mirror.
Focal Length:
The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’
The focal length is basically equal to the half of the radius of curvature of the mirror:
[tex]f = \frac{r}{2}[/tex]
where,
f = focal length = ?
r = radius of curvature = 2.51 cm
[tex]f = \frac{2.51\ cm}{2}[/tex]
f = 1.255 cm
Which describes the greenhouse effect?
a. an artificial process
b. a dangerous process
c. a natural process
d. new process
c. a natural process
It is a natural process