Answer:
The effective dose is [tex]E = 11.1\ mS_v[/tex]
Explanation:
From the question we are told that
The whole-body does received from 100-keV neutrons is [tex]D_1 = 0.30 mGy = 0.30 *10^{-3} \ Gy[/tex]
The whole-body does received from 1.5-MeV neutrons is [tex]D_2 = 0.19 mGy = 0.19 *10^{-3} \ Gy[/tex]
The whole-body does received from gamma rays is [tex]D_3= 4.3 mGy = 4.3 *10^{-3} \ Gy[/tex]
Generally from the table of radiation weight factor ,
The radiation weight factor for 100-keV neutrons is [tex]R_1 = 10[/tex]
The radiation weight factor for 100-keV neutrons is [tex]R_2 = 20[/tex]
The radiation weight factor for gamma rays is [tex]R_3 = 1[/tex]
Generally the effective does is mathematically represented as
[tex]E = R_1 * D_1 + R_2 * D_2 + R_3 * D_3[/tex]
=> [tex]E = 10 * 0.30 +20 * 0.19 +1 * 4.3[/tex]
=> [tex]E = 11.1\ mS_v[/tex]
A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]
A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?
Answer:
The maximum angular velocity is 20 rad/s
Explanation:
Given;
torque, τ = 10 N
maximum mechanical power, P = 200 J/s
The output power of the pmdc is given as;
P = τω
where;
P is the maximum mechanical power
ω is the maximum angular velocity
ω = P / τ
ω = (200) / (10)
ω = 20 rad/s
Therefore, the maximum angular velocity is 20 rad/s
Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container. The liquids have refractive indices of 1.75 and 1.33. What is the apparent depth of the combination when viewed from above?
Answer:
The apparent depth d = 19.8495 cm
Explanation:
The equation for apparent depth can be expressed as:
[tex]d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}[/tex]
here;
[tex]d_1 = d_2 = 15 \ cm[/tex]
[tex]\mu_1[/tex] = refractive index in the first liquid = 1.75
[tex]\mu_2[/tex] = refractive index in the second liqquid= 1.33
∴
[tex]d = \dfrac{15}{1.75}+\dfrac{15}{1.33}[/tex]
[tex]d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})[/tex]
[tex]d = 15( 0.5714 +0.7519)[/tex]
d = 15(1.3233 ) cm
d = 19.8495 cm
WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS?
Explain
Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.
Explanation:
Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)
Which statement describes an example of static electricity?
A. When a light switch is flipped, electrons flow to a lightbulb.
B. Wind turns a generator, causing electrons to flow into power lines.
C. Electrons move from one end of a copper wire to another.
D. Positively and negatively charged particles are attracted to each other.
The answer is Positively and negatively charged particles are attracted to each other.
Which equation is most likely used to determine the acceleration from a velocity vs. time graph?
Answer:
a = t over delta v.
Explanation:
The amplitude of a pendulum is doubled. This means:
a
the pendulum will have twice its original mass.
b
the frequency of the pendulum will be twice as high.
c
the pendulum will swing twice as far away from the center.
d
the period of the pendulum will be twice as long.
Answer:
the period of the pendulum will be twice as long.
Explanation:
because i looked it up
A 2.80 kg mass is dropped from a height of 4.50 m. Find its potential energy when it reaches the ground.
Answer:
123.48J
Explanation:
Given parameters:
Mass of the ball = 2.8kg
Height = 4.5m
Unknown:
Potential energy = ?
Solution:
The potential energy is the energy due to the position of a body. It is mathematically given as;
P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the parameters and solve;
P.E = 2.8 x 4.5 x 9.8 = 123.48J
Answer:
0
Explanation:
There is 0 PE when its on the ground
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
a cyclist accelerates at a rate of 7.0 m/s2. how long will it take the cyclist to go from a velocity of 4 m/s to a velocity of 18 m/s?
Answer:
2.57 seconds (rounded to 2.6 Seconds)
Step-by-step explanation:
Great question, it is always good to ask away and get rid of any doubts that you may be having.
Before we can solve this question we need to create a formula that calculates the final speed. The formula will be the following,
Where:
Vf is the final Velocity
Vi is the initial velocity
A is the acceleration
t is the time in seconds
Now that we have the formula we can plug in the values given to us in the question and solve for the amount of time (t).
Finally, we can see that it would take the cyclist 2.57 seconds (approximately) to reach a speed of 18 m/s
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
When will the Social Security fund dry up at its current level?
A.
2022
B.
2027
C.
2042
D.
2097
E.
2117
Answer:
C. 2042
Explanation:
trust me I know the answer
In 2042 the Social Security fund will dry up at its current level. The correct option is C.
The answer is not entirely clear, as projections can vary depending on various economic and demographic factors. However, based on the 2021 Social Security Trustees Report, the projected depletion date for the Social Security Trust Fund is 2034. This means that at the current rate of payouts and contributions, the Trust Fund will only be able to pay out about 78% of scheduled benefits after 2034.
Therefore, the closest option to the projected depletion date of the Social Security Trust Fund is C. 2042. However, it is important to note that this projection may change as economic and demographic factors change over time.
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A book is sitting on a table. Which of the following is true about the table?
Answer:
Its pulling down on the book
Explanation: if it was pushing up the book would be floating and the other choices don't make sense
I need help ASAP!
1. Two wave pulses move towards each other as shown below. The pulses have the same width and amplitudes.
What is the resulting wave pattern when the centers of the two pulses meet?
Answer:
b.
Explanation:
The resulting wave pattern when the centers of the two pulses meet is undergo constructive interference.
When two waves meet?Waves are pulses of energy that propagate through space periodically. When two waves overlap in the same region of space, interference occurs, which results in another wave with different intensity. These variations in the intensity of the resulting wave are called interference fringes.
The two pulses propagate with the same phase and in opposite directions, when they meet, they suffer constructive interference, which will cause the sum of the amplitudes. After interference, each wave goes its way as if nothing happened.
See more about waves at brainly.com/question/3639648
Can I get help on this question please
it would be the 3rd one. so C
A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide
Answer:
[tex]\theta = 16.70 ^{\circ}[/tex]
Explanation:
The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.
[tex]\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta[/tex]Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.
Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.
[tex]w=mg[/tex]We are given the mass of the block (kg) and we know that g = 9.8 m/s².
[tex]w=(5)(9.8) = 49 \ \text{N}[/tex]Now we can use this force in the equation:
[tex]\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}[/tex]Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.
[tex]\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}[/tex] [tex]0.30=\text{tan} \theta[/tex]Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].
Evaluate this equation by taking the inverse tangent of both sides of the equation.
[tex]\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)[/tex] [tex]\text{tan}^-^1 (0.30) =\theta[/tex] [tex]\theta = 16.69924423[/tex]The minimum angle at which the block will begin to slide is about 16.70 degrees.
A 950 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72 m, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?
Compute the car's weight:
W = m g = (950 kg) (9.8 m/s²) = 9310 N
The net vertical force on the car is
∑ F = N - W = 0
so the normal force has magnitude
N = W = 9310 N
Then the friction force that keeps the car from skidding has magnitude F = µ N, where µ is the coefficient of friction, and it's friction that makes up the net horizontal force on the car. By Newton's second law, we have
F = m a
µ N = m v ² / R
µ (9310 N) = (950 kg) (25 m/s)² / (72 m)
µ ≈ 0.89
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
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which factor does not affect the strength of an electromagnet
Answer:
the placement of the ammeter in the circuit
Explanation:
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of
20
N
20N20, start text, N, end text for
15.0
m
15.0m15, point, 0, start text, m, end text.
How much kinetic energy does the dirt gain?
Answer:
300 JoulesExplanation:
This is a common question on Khan Academy's "Calculating change in kinetic energy from a force" practice exercises. (AP Physics 1)
The simplest method to use is the following: [tex]W = F * d * cos(theta)[/tex], where W represents work (joules), F represents force (newtons), d represents distance (meters), and theta represents the angle of the force that's being applied.In this scenario, the force being applied is horizontal, so we can remove the [tex]cos(theta)[/tex] from our equation.So, our equation is now: [tex]W = F * d[/tex]. This would mean that [tex]W = 20 * 15[/tex], which is equal to [tex]300[/tex]. Our answer is 300 joules. (this value is positive and not negative because kinetic energy is being GAINED, not LOST)Here's the real question without all the formatting:
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20 N for 15.0 m. How much kinetic energy does the dirt gain?A truck pushes a pile of dirt in the horizontal direction with a force of 20 N to a distance of 15 m then the kinetic energy of the dirt is 300 J.
What is work?In physics, the term "work" refers to the measurement of energy transfer that takes place when an item is moved over a distance by an externally applied, at least a portion of which is applied inside the direction of the displacement.
The duration of the path is multiplied by the component of a force acting along the path to calculate work if the forces are constant. The work W is theoretically equivalent to the pressure f times the range d, or W = fd, to represent this idea. Work is done when a force is applied at an angle of to a displacement, or W = fd cos.
The work done, [tex]W = Force * Displacement[/tex]
W = 15× 20
W = 300 J
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If I can travel 20m in 18 seconds how far can I go in 10 minutes?
Answer:
36 s
Explanation:
20 m = 18 s
10 m = ?
20 × 18 = 360÷ 10
= 36 sec
The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
Putting the selected answers in parenthesis
A cation has a (Negative, Positive, Neutral) charge, because they have (Lost, Maintained, Gained) electrons. An anion has a (Negative, Positive, Neutral) charge, because they have (Lost, Maintained, Gained) electrons. An atom with the same number and protons and neutrons have a (Negative, Positive, Neutral) charge.
Answer:
gimmie
Explanation:
20 pts
Which of the following statements is true?
1..LIghtning is a form of statlc energy.
2..Water Is a conductor of electricIty.
3Electricity must flow in a complete circuit.
4 all of the above
Answer: 2
Explanation:
:}
What can you infer from the fact that metals are good conductors of electricity?
Answer:
Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.
10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)
Answer: mass = 48.47 kg.
Explanation:
Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .
Given: Weight = 475 N
[tex]g= 9.8\ m/s^2[/tex]
Substitute all values in formula , we get
[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]
Hence, his mass = 48.47 kg.
to determine the height of a steep cliff an experimenter stations a sensor on the top of the cliff then fires a pellet vertically upward with an initial velocity of 80 m/s . the sensor reports that the pellet reached a maximum height 3 meters above the edge of the cliff. how high is the cliff?
a. 77 m
b. 237 m
c. 317 m
d. 637 m
e. 797 m
Answer:
c. 317 m
Explanation:
Vertical Launch Upwards
It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]
We'll assume the acceleration of gravity as [tex]g=10\ m/s^2[/tex].The pellet is vertically upward launched with vo=80 m/s. The maximum height is:
[tex]\displaystyle h_m=\frac{80^2}{2*10}=320[/tex]
[tex]h_m = 320\ m[/tex]
That height is 3 meters above the edge of the cliff, thus the cliff is 320-3=317 m hight
c. 317 m
Two satellites orbit the same planet. if a satellite A has an orbit radius r and satellite B has an orbit radius 2r, find the ratio of the period of satellite B to the period of satellite A.
Given :
Two satellites orbit the same planet. if a satellite A has an orbit radius r and satellite B has an orbit radius 2r.
To Find :
The ratio of the period of satellite B to the period of satellite A.
Solution :
We know, square of time period (T) is directly proportional to cube root of radius (r) .
So,
[tex]\dfrac{T_B^2}{T_A^2}=\dfrac{(2r)^3}{r^3}\\\\\dfrac{T_B}{T_A}=2\sqrt{2}[/tex]
Therefore, the ratio of the period of satellite B to the period of satellite A is 2√2 .
Show that the pressure in liquids is given by density multiplied by gravity multiplied by height
Answer
The answer to your question is given below.
Explanation:
Pressure (P) = force (F) / Area (A)
P = F/A ........ (1)
Recall:
1. Force (F) = mass (m) × acceleration due to gravity (g)
F = mg
2. Volume (V) = Area (A) × Height (h)
V = Ah
Divide both side by h
A = V/h
Substitute the value of F and A into equation 1.
P = F/A
P = mg ÷ V/h
P = mg × h/V
P = mgh/V.... (2)
Recall:
Density (d) = mass (m) /volume (V)
d = m/V
Replace m/V in equation (2) with d.
P = mgh/V
P = dgh
Where:
P is the pressure.
d is the density.
g is acceleration due to gravity.
h is height.
Pressure = Density × gravity × height
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
True or False: As water evaporates, it removes a lot of heat with it.
Answer:
True, when it evaporates heat goes down, because it’s up the the sky.
Facts: In order for water to evaporate, hydrogen bonds must be broken. Water's heat of vaporization is 540 cal/g.
Because of the need for so much energy to evaporate, as water leaves the surface from which it is evaporating and removes a lot of heat with it.