Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s
You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:
Answer:
18
Explanation:
I'm pretty sure I got it right
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?
Answer:
9.89 m/s.
Explanation:
Given that,
The radius of the circular arc, r = 25 m
The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²
Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :
[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]
So, the maximum speed of the car should be 9.89 m/s.
Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength.
Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10−4 m2, a typical value.)
Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area?
Answer:
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
Explanation:
Given:
cross sectional area of the bone, [tex]a=4.8 \times 10^{-4} ~m^2[/tex]
factor of up-scaling the dimensions, [tex]s=10[/tex]
Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.
The scaled up area is:
[tex]a_s=a\times s^2[/tex]
[tex]a_s=[4.8 \times 10^{-4}]\times 10^2[/tex]
[tex]a_s=4.8\times 10^{-2}~m^2[/tex]
The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. The new cross-section area will be 4.8×10⁻² m².
What is the area?The area is defined as the space covered by an object in 2 d dimension. For a rectangle, it is a product of length and breadth. Its unit is m².
Given data in the problem
a is the crossectional area of conical bone = 4.8×10⁻⁴m².
s is the factor of up-scaling the dimensions =10
For two degrees of dimension, the upscaled area will be square of the given area.
The scaled-up area will be
[tex]\rm a_s=a\times s^2\\\\ a_s= 4.8\times10^{-4}\times {10}^2\\\\\ \rm a_s=4.8\times10^{-2}\;m^2[/tex]
Hence the new cross-section area will be 4.8×10⁻² m².
To learn more about the area refer to the link;
https://brainly.com/question/1631786
The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Answer:
The potential at a distance of 0.9 m is 266.67 V.
Explanation:
Charge = Q
Potential is 400 V at a distance 0.6 m .
Let the potential is V at a distance 0.9 m.
Use the formula of potential.
[tex]V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V[/tex]
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm
What is (a) the x component and (b) the y component of the net electric field at the square's center
Answer:
What is (a) the x component and (b) the y component of the net electric field at the square's center
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6. 7 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 3.0 m from the center of the circle.
Answer:
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]
Explanation:
The centripetal acceleration is given as follows:
[tex]a_c = \frac{v^2}{r}\\[/tex]
where,
ac = centripetal acceleration
v = linear speed = rω
r = radius
ω = angular speed
Therefore,
[tex]a_c = \frac{(r\omega)^2}{r}\\\\a_c = r\omega^2[/tex]
Therefore, the ratio will be:
[tex]\frac{a_{c1}}{a_{c2}} = \frac{r_1\omega^2}{r_2\omega^2}\\\\\frac{a_{c1}}{a_{c2}} = \frac{r_1}{r_2}\\\\[/tex]
where,
r₁ = 6.7 m
r₂ = 3 m
Therefore,
[tex]\frac{a_{c1}}{a_{c2}} = \frac{6.7\ m}{3\ m}\\\\[/tex]
[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]
During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race
Answer:
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Explanation:
Given;
initial distance covered by Bolt, d = 200 m
time of this motion, t = 19.3 s
The second distance covered by Bolt, = 100 m
Assuming Bolt maintained the same acceleration for both races.
His acceleration can be determined from the 200 m race.
d = ut + ¹/₂at²
where;
u is his initial velocity = 0
d = ¹/₂at²
[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]
Let the final or maximum velocity for the 100 m race = v
v² = u² + 2ad₂
v² = 2 x 1.074 x 100
v² = 214.8
v = √214.8
v = 14.66 m/s
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
https://brainly.com/question/9410676
A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
What happens to the acceleration if you triple the force that you apply to the painting with your hand? (Use the values from the example given in the previous part of the lecture.) Submit All Answers Answer: Not yet correct, tries 1/5 3. A driver slams on the car brakes, and the car skids to a halt. Which of the free body diagrams below best matches the braking force on the car. (Note: The car is moving in the forward direction to the right.] (A) (B) (C) (D) No more tries. Hint: (Explanation) The answer is A. The car is moving to the right and slowing down, so the acceleration points to the left. The only significant force acting on the car is the braking force, so this must be pointing left because the net force always shares the same direction as the object's acceleration. 4. Suppose that the car comes to a stop from a speed of 40 mi/hr in 24 seconds. What was the car's acceleration rate (assuming it is constant). Answer: Submit Al Answers Last Answer: 55 N Only a number required, Computer reads units of N, tries 0/5. 5. What is the magnitude (or strength) of the braking force acting on the car? [The car's mass is 1200 kg.) Answer: Submit Al Answers Last Answer: 55N Not yet correct, tries 0/5
Answer:
2) when acceleration triples force triples, 3) a diagram with dynamic friction force in the opposite direction of movement of the car
4) a = 2.44 ft / s², 5) fr = 894.3 N
Explanation:
In this exercise you are asked to answer some short questions
2) Newton's second law is
F = m a
when acceleration triples force triples
3) Unfortunately, the diagrams are not shown, but the correct one is one where the axis of movement has a friction force in the opposite direction of movement, as well as indicating that the car slips, the friction coefficient of dynamic.
The correct answer is: a diagram with dynamic friction force in the opposite direction of movement of the car
4) let's use the scientific expressions
v = v₀ - a t
as the car stops v = 0
a = v₀ / t
let's reduce the magnitudes
v₀ = 40 mile / h ([tex]\frac{5280 ft}{1 mile}[/tex]) ([tex]\frac{1 h}{3600 s}[/tex]) = 58.667 ft / s
a = 58.667 / 24
a = 2.44 ft / s²
5) let's use Newton's second law
fr = m a
We must be careful not to mix the units, we will reduce the acceleration to the system Yes
a = 2.44 ft / s² (1 m / 3.28 ft) = 0.745 m / s²
fr = 1200 0.745
fr = 894.3 N
Please assist with solving this problem and showing the steps
Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s
Answer:
1.43 s
Explanation:
Applying,
a = (v-u)/t........... Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (v-u)/a........... Equation 2
From the question,
Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²
Substitute into equation 2
t = (2-0)/1.4
t = 1.43 s
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)
Answer:
................ftf6x
The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?
Explanation:
The period T of a simple pendulum is given by
[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]
Doubling the length of the pendulum gives us a new period T'
[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
Where is the sun in relation to earths orbit
I HAVE A PHYSICS TEST, ITS 25 QUESTIONS AND I HAVE ABOUT AN HOUR TO SOLVE IT PLEASE IF YOU'RE GOOD AT PHYSICS CONTACT ME ASAP
Answer:
yes sir
Explanation:
A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at a constant vertical acceleration of 6.89 m/s2 the initial plane horizontal speed was 86.9 m/s, how far away from the survivor will it hit the waves
Answer: 19.15 meters on the waves away from the survivor.
Explanation:
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
