Answer:
The tempature of the matter increases.
Explanation:
The matter is heated by friction, weather that is through the air or if its touching the ground like a car.
please mark me as brainliest of this is right, thanks
uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J
Answer:
a) W = - 1.752 10⁻¹⁸ J, b) U = + 1.752 10⁻¹⁸ J
Explanation:
a) work is defined by
W = F . x
the bold letters indicate vectors, in this case the force is electric
F = q E
we substitute
F = q E x
the charge of the electron is
q = - e
F = - e E x
let's calculate
W = - 1.6 10⁻¹⁹ 365 3 10⁻²
W = - 1.752 10⁻¹⁸ J
b) the change in potential energy is
U = q ΔV
the potential difference is
ΔV = - E. Δs
we substitute
U = - q E Δs
the charge of the electron is
q = - e
U = e E Δs
we calculate
U = 1.6 10⁻¹⁹ 365 3 10⁻²
U = + 1.752 10⁻¹⁸ J
A hockey ball is flicked of the ground with initial velocity of 2.0m/s upwards and 10m/s horizontally. Calculate the distance travelled from the point where the ball is flicked and to the point where the ball hits the ground.
Answer:
imma try and fail again and again
Explain how the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure. Sun Earth at position B Earth at position A
Answer:
The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.
An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
[tex]p.d' = 37.6 V[/tex]
Explanation:
From the question we are told that:
Potential difference [tex]p.d=18.8V[/tex]
New Capacitor [tex]C_1=C_2/2[/tex]
Generally the equation for Capacitor capacitance is mathematically given by
[tex]C=\frac{eA}{d}[/tex]
Generally the equation for New p.d' is mathematically given by
[tex]C_2V=C_1*p.d'[/tex]
[tex]p.d' = 2V[/tex]
[tex]p.d'= 2 * 18.8[/tex]
[tex]p.d' = 37.6 V[/tex]
5. For the speaker in this circuit, the voltage across it is always proportional to the current through it. Find the maximum amount of power that the circuit can deliver to the speaker.
Answer:
speaker64
--------
34x
Explanation:
64-34
x
speaker
4
2
4
788
- circuit
voltage
100000
x.34
Sorry but you have no picture shown
An infinite plane lies in the yz-plane and it has a uniform surface charge density.
The electric field at a distance x from the plane
a.) decreases as 1/x^2
b.) increases linearly with x
c.) is undertermined
d.) decreases linearly with x
e.) is constant and does not depend on x
Answer:
So the correct answer is letter e)
Explanation:
The electric field of an infinite yz-plane with a uniform surface charge density (σ) is given by:
[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]
Where ε₀ is the electric permitivity.
As we see, this electric field does not depend on distance, so the correct answer is letter e)
I hope it helps you!
b. The stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. . 3 At what distance x does the stream strike the floor?
Answer:
34.64 cm
Explanation:
Given that:
The depth of the hole h = 10 cm
height of the water holding in the tank H = 40 cm
For a stream of flowing water, the distance (x) at which the stream strikes the floor can be computed by using the formula;
[tex]x = 2 \sqrt{h(H-h)}[/tex]
[tex]x = 2 \sqrt{10(40-10)}[/tex]
[tex]x = 2 \sqrt{10(30)}[/tex]
[tex]x = 2 \sqrt{300}[/tex]
[tex]x = 2 \times 17.32[/tex]
x = 34.64 cm
True or false: Increasing the Young’s modulus of a beam in bending will cause it to deflect less.
Answer:
false?
Explanation:
The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.
Answer:
I think the answer is False.
What distance do I cover if I travel 10 m E, then 6 mW, then 12 m E?
A. 16 m
B. 28 m
C. 16 m E
D. 28 m E
Answer:
C. 16 m E
Explanation:
Applying,
The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted
From the question above,
Step 1: Total distance covered towards east = 10+12 = 22 m E
Step2: Total distance covered towards west = 6 m W
Therefore, the resultant distance traveled = 22-6 = 16 m E
Hence the right option is C. 16 m E
The mass is released from the top of the incline and slides down the incline. The maximum velocity (taken the instant before the mass reaches the bottom of the incline) is 1.06 m/s. What is the kinetic energy at that time
Answer:
0.28 J
Explanation:
Let the mass of the object is 0.5 kg
The maximum velocity of the object is 1.06 m/s.
We need to find the kinetic energy at that time. It is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]
So, the required kinetic energy is equal to 0.28 J.
3) A fahrenheit thermometer shaded from the sun on a hot day reads 101⁰ what is the temperature on the centigrade scale?
Answer:
38.33°C
Explanation:
Applying,
180/100 = (F-32)/C............. Equation 1
Where F = Temperature of the hot day in fehrenheit, C = Temperature of the hot day in centigrade.
make C the subject of the equation
C = 100(F-32)/180.............. Equation 2
From the question,
Given: F = 101°F
Substitute into equation 2
C = 100(101-32)/180
C = 38.33°C
A ball rolled along a horizontal surface comes to rest in a distance of 72m in 6s. Its initial velocity
and deceleration are
and
Answer:
1. Initial velocity = 24 m/s
2. Deceleration = –4 m/s²
Explanation:
From the question given above, the following data were obtained:
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
1. Determination of the initial velocity.
Distance travelled (s) = 72 m
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) =?
s = (u + v)t / 2
72 = (u + 0) × 6 / 2
72 = u × 3
Divide both side by 3
u = 72 / 3
u = 24 m/s
2. Determination of the deceleration.
Time (t) = 6 s
Final velocity (v) = 0 m/s
Initial velocity (u) = 24 m/s
Deceleration (a) =?
v = u + at
0 = 24 + (a × 6)
0 = 24 + 6a
Collect like terms
0 – 24 = 6a
–24 = 6a
Divide both side by 6
a = –24 / 6
a = –4 m/s²
write physical quantities and its unit
length= metre
mass= kg
time= second
temperature = kelvin
current= ampere
luminous intensity= candela
Amount of substance = mole
etc
I hope this will help you
stay safe
Cho điện cực dưới điện cực trên . Hàm thế biến thiên theo qui luật:
Xác định sự phân bố điện tích khối và .
Answer:
Quantum theory gives the concept of
What is the net force on the side of the container The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Answer:
26467.21 N
Explanation:
Initial height of water ( h1 ) = 5 m
diameter of container ( d1 )= 100 cm
pressure inside the container ( p1 )= 0.850 atm
Diameter of faucet ( d2 )= 1 inch
Calculate the value of the net force on the side of container
lets assume ; pressure outside the container ( p2 ) = 1 atm
Fnet = ( P1A1 + mg ) - ( P2A2 )
= [ ( 0.85 * 101325 ) ( π(1/2)^2 ) + mg ) - [ ( 101325 )( π )(0.0127)^2 ]
= [ 16902.2766 + pvg ] - [ 51.3161 ]
where ; pvg = pAhg = 1000 * π ( 1/2 )^2 * 5 * 9.8 = 9616.25
= [ 16902.2766 + 9616.25 ] - [ 51.3161 ] = 26467.21 N ( downwards )
A pulse traveled the length of a stretched spring the pulse transferred...A)energy only B)mass only C)both energy and mass D) neither energy nor mass
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.
The human eye can readily detect wavelengths from about 400 nm to 700 nm. Part A If white light illuminates a diffraction grating having 910 lines/mm , over what range of angles does the visible m
Answer:
The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".
Explanation:
According to the question,
Slit width,
[tex]d=\frac{1}{910 \ lines/mm}[/tex]
[tex]=\frac{1}{910\times 10^3}[/tex]
[tex]=1.099\times 10^{-6} \ m[/tex]
The condition far first order maxima will be:
⇒ [tex]d Sin \theta = 1 \lambda[/tex]
Now,
⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=21.344^{\circ}[/tex]
⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=39.56^{\circ}[/tex]
Place each description under the correct theory
Gravity is an attractive force.
Universal Law of Gravitation
General Theory of Relativity
Mass and distance affect force.
Time and space are absolute,
Time and space are relative.
Gravity is due to space-time curving.
Mass affects space-time curving.
Answer:
1) Law of Universal Gravitation Gravity is an attractive force
5) General relativity Gravity is due to the curvature of spacetime
Explanation:
In this exercise you are asked to relate the correct theory and its explanation
Theory Explanation
1) Law of Universal Gravitation Gravity is an attractive force
2) Law of universal gravitation Mass and distance affect force
3) Classical mechanics time and space are absolute
4) Special relativity Time and space are relative
5) General relativity Gravity is due to the curvature of
spacetime
6) General relativity Mass affects the curvature of space - time
Answer:
Explanation:
edge2022
Explain how solar radiation is converted into thermal energy
Answer:
See the explanation below
Explanation:
Explain how solar radiation is converted into thermal energy
From the concept of the particulate nature of matter, when the gases(air) receives a sufficient amount of radiation from the sun they tend to move in a constant rapid random motion as they move they hit one another and hence heat in form of thermal energy
A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{\rm total}=\Delta K[/tex]
or
[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]
Compute the work performed by friction:
[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]
[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]
[tex]\implies \boxed{\mu\approx0.45}[/tex]
The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
What is coefficient of friction?Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
[tex]W_{total}=\Delta K[/tex]
or
[tex]W_{friction}+W_{spring}=0-K=-K[/tex]
where K is the block's kinetic energy at the equilibrium point,
[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]
Compute the work performed by friction:
[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]
By Newton's second law, the net vertical force on the block is
∑ F = n - mg = 0 ==> n = mg
where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.
So we have
[tex]W_{friction}=-f(0.20)[/tex]
[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]
[tex]\mu =0.45[/tex]
Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]
To know more about Coefficient of friction follow
https://brainly.com/question/136431
4
Type the correct answer in the box. Use numerals instead of words.
Anne has a sample of a substance. Its volume is 20 cm and its mass is 100 grams. What is the sample's density?
The sample's density is
g/cm?
Reset
Next
Answer:
5g/cm
Explanation:
denisty=mass/volume
100/20
5g/cm
Galaxies that are 400 million light years away have a red shift of 0.03 approximately. A radio wave coming from one of these galaxies has an observed wavelength of 125 meters. What is the emitted wavelength in meters and the observed frequency in Megahertz
Answer:
The correct answer is "121.36 meters and 2.40 MHz".
Explanation:
Given:
Red shift,
[tex]\frac{v}{c}=\frac{\lambda - \lambda_0}{\lambda_0} = 0.03[/tex]
Wavelength,
[tex]\lambda = 125 \ meters[/tex]
The observed frequency will be:
⇒ [tex]f = \frac{c}{\lambda}[/tex]
[tex]=\frac{3\times 10^8}{125}[/tex]
[tex]=24\times 10^5[/tex]
[tex]= 2.40\times 10^6[/tex]
[tex]=2.40 \ MHz[/tex]
hence,
The Emitted wavelength will be:
⇒ [tex]\frac{125-\lambda_0}{\lambda_0} =0.03[/tex]
[tex]\frac{125}{\lambda_0}-1 =0.03[/tex]
[tex]125=1.03 \ \lambda_0[/tex]
[tex]\lambda_0=\frac{125}{1.03}[/tex]
[tex]=121.36 \ m[/tex]
A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?
Answer:
[tex]t=0.50cm[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lamda=3c[/tex]m
Refraction Index [tex]n=1.50[/tex]
Generally the equation for Destructive interference for Normal incidence is mathematically given by
[tex]2nt=m(\frac{1}{2})\lambda[/tex]
Since Minimum Thickness occurs at
At [tex]m=0[/tex]
Therefore
[tex]t=\frac{\lambda}{2}[/tex]
[tex]t=\frac{3}{4(1.50)}[/tex]
[tex]t=0.50cm[/tex]
5. A big wheel has a diameter of 5 m and a mass of 1500 kg when fully laden with people. a) Work out the moment of inertia of the big wheel. (Hint: which shape from the ones given on p114 would be most suitable? b) When the wheel is rotating at full speed, a person has a linear velocity of 3 m/s. What is the angular velocity of this person? c) What is the rotational kinetic energy at this speed? d) A motor takes 10 seconds to accelerate the wheel from rest to a linear velocity on the circumference of 3 m/s. What is the power of the motor?
Answer:
a) I = 3.75 10⁴ kg m², b) w = 0.6 rad / s, c) K = 6.75 10³ J, d) P = 6.75 10² W
Explanation:
This is a rotations exercise
a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.
We are going to approximate the reda with a thin ring
I = M r²
I = 1500 5²
I = 3.75 10⁴ kg m²
b) angular and linear velocity are related
v = w r
w = v / r
w = 3/5
w = 0.6 rad / s
c) the expression for kinetic energy is
K = ½ I w²
K = ½ 3.75 10⁴ 0.6²
K = 6.75 10³ J
d) the power is
P = W / t
to find the work we use the relationship between work and the variation of kinetic energy
W = ΔK = K_f - K₀
the system part of rest wo = 0
W = K_f
W = 6.75 10³ J
we calculate
P = 6.75 10³/10
P = 6.75 10² W
The thermal efficiency (in %) of a system that undergoes a power cycle while receiving 1000 kJ of energy by heat transfer from a hot reservoir at 1000 K and discharging 500 kJ of energy by heat transfer to a cold reservoir at 400 K is:
Answer:
η = 0.5 = 50%
Explanation:
The efficiency of the power cycle is given by the following formula:
[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]
where,
where,
η = efficiency = ?
Q₁ = heat received from hot reservoir = 1000 KJ
Q₂ = heat discharged to cold reservoir = 500 KJ
Therefore,
[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]
η = 0.5 = 50%
In a robotics circuit, a voltage source of 75V is supplying a current, I to a series circuit of 5
resistances. Resistance, R1 = 5 KΩ and R2 = 10 KΩ. The voltage drops across 3 black boxes of
resistances R3 , R4 and R5 are 15V, 20V and 25V respectively. The current through the black
box of resistance, R5 is measured as 1mA. Calculate the voltage V1 and V2 across the
resistance R1 and R2 using the Voltage Divider Rule.
Answer:
In the given circuit, R
2
,R
6
and R
4
are in series. So,
R
1
′
=7+5+12=24Ω
Now R
1
′
and R
5
′
are in parallel. So,
R
2
′
1
=
8
1
+
24
1
=
24
3+1
=
24
4
=
6
1
R
2
′
=6ohm.
Now R
2
′
,R
1
and R
3
are in series. So,
R=R
2
′
+R
1
+R
3
=6+3+2=11ohm.
We know i=
R+r
E
=
11+1
6
=
12
6
=
2
1
i=0.5amp.
Consult Multiple-Concept Example 11 for background material relating to this problem. A small rubber wheel on the shaft of a bicycle generator presses against the bike tire and turns the coil of the generator at an angular speed that is 40 times as great as the angular speed of the tire itself. Each tire has a radius of 0.330 m. The coil consists of 163 turns, has an area of 3.23 x 10-3 m2, and rotates in a 0.0948-T magnetic field. The bicycle starts from rest and has an acceleration of 0.601 m/s2. What is the peak emf produced by the generator at the end of 6.72 s
Answer:
[tex]Emf=24.4V[/tex]
Explanation:
From the question we are told that:
Angular speed [tex]\omega=40*\omega'[/tex]
Radius [tex]r=0.33m[/tex]
No. Turns [tex]N=163[/tex]
Area [tex]A= 3.23 * 10^{-3} m^2[/tex]
Magnetic field. [tex]B=0.0948T[/tex]
Acceleration [tex]a=0.60m/s^2[/tex]
Time [tex]t=6.72s[/tex]
Generally the equation for momentum is mathematically given by
[tex]\omega'=\omega_o+\frac[a}{r}t[/tex]
[tex]\omega'=0+\frac{0.60}{0.33}*6.72[/tex]
[tex]W=12.22rad/s^2[/tex]
Therefore
[tex]\omega=Aw[/tex]
[tex]\omega=12.22*40[/tex]
[tex]\omega=488.7rads/s[/tex]
Generally the equation for Peak emf is mathematically given by
[tex]Emf=NBA \omega[/tex]
[tex]Emf=163*0.0948* 3.23 * 10^{-3} m^2*488.7rads/s[/tex]
[tex]Emf=24.4V[/tex]
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm. Fwithout belt
This question is incomplete, the complete question is;
Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.
1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?
Fwith belt =
2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.
Fwithout belt =
Answer:
1) The Net force on the driver with seat belt is 10.3 KN
2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
Explanation:
Given the data in the question;
from the equation of motion, v² = u² + 2as
we solve for a
a = (v² - u²)/2s ----- let this be equation 1
we know that, F = ma ------- let this be equation 2
so from equation 1 and 2
F = m( (v² - u²)/2s )
where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.
1)
Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.
i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m
so we substitute the given values into the equation;
F = 70( ((0)² - (18)²) / 2 × 1.1 )
F = 70 × ( -324 / 2.4 )
F = 70 × -147.2727
F = -10309.09 N
F = -10.3 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwith belt = 10.3 KN
Therefore, Net force of the driver is 10.3 KN
2)
No sit belt,
m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m
we substitute
F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )
F = 70 × ( -324 / 0.022 )
F = 70 × -14727.2727
F = -1030909.08 N
F = -1030.9 KN
The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.
Fwithout belt = 1030.9 KN
Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN
If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the track. Assume the track is frictionless. Hint: use the definitions of acceleration and Newton’s second law.
Answer:
hehe
Explanation:
I dont know because I am a noob ant study
3. If you change the resistance of the resistor:
a. How does the current through the circuit change? (answer, explain, evidence)
b. How does the voltage of the battery change? (answer, explain, evidence)
Answer:
Explanation:
Changing the resistance of a resistor means the resistance is either increased or decreased.
a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.
Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;
V = IR
6 = I x 30
I = 0.2 A
If the resistance is changed to 50 Ohm's, then:
I = 0.12 A
(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.
In the previous example, if the resistance is changed to 5 Ohm's, then:
V = IR
6 = I x 5
I = 1.2 A
(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.
