Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
What formula could be used to find distance if you know the speed an the time
Answer: d = st
Explanation:
We know that the distance is equal to the rate (speed) times the time
d = st
A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?
Answer:
The height of the hill is 0.46 m.
Explanation:
Given;
mass of the child and sled, m = 50 kg
initial velocity of the sled, u = 0
final velocity of the sled, v = 3 m/s
The height of the high is calculated from the law of conservation of energy;
P.E at top = K.E at bottom
mgh = ¹/₂mv²
gh = ¹/₂v²
[tex]h = \frac{v^2}{2g} \\\\h = \frac{3^2}{2\times 9.8} \\\\h = 0.46 \ m[/tex]
Therefore, the height of the hill is 0.46 m.
Can you share me your answers ❤️❤️
Answer:
Depending, on how much it's push against together.
Explanation:
Since, the two objects are getting in contact. But, if it's a type of item/thing there's a different frictions, but I know it's normal friction when two objects comes in contact. But, its depending on how much you push it against the two items.
Which result is more accurate: the slope or the mean value
What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds
The time required for one complete cycle of a mass oscillating at the end of a spring is 0.40 s. What is the frequency of oscillation?
Answer:
the frequency of the oscillation is 2.5 Hz.
Explanation:
Given;
time to complete the oscillation, t = 0.4 s
number of oscillations, n = 1
The frequency of the oscillation is calculated as;
[tex]F = \frac{n}{t} \\\\F = \frac{1}{0.4} \\\\F = 2.5 \ Hz[/tex]
Therefore, the frequency of the oscillation is 2.5 Hz.
Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organism does this most likely belong to?
A. A plant B. An animal C. A fungus D. A bacterium
Answer:
A plant
Explanation:
because animals don't have cell walls, and fungus and bacteria dont have chloroplasts
The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters before coming to rest in a lock containing 75,000 cubic meters of fresh water. The specific heat of water is 4200 joules per kilogram degree Celsius. Assume all energy is transferred from the ship to the water. Determine the change in temperature of the water in degrees Celsius .
Answer:
ΔT = 17.11 °C
Explanation:
In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.
At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.
So, let's calculate first the potential energy of the ship:
E = mgh (1)
We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:
E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m
E = 5.39x10⁹ J
Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:
E = m * C * ΔT (2)
We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:
ΔT = E / m * C (3)
ΔT = 5.39x10⁹ / 4200 * 75000
ΔT = 17.11 °CHope this helps
A spring is stretched 5 mm by a force of 125 N. How much will the spring stretch
when 250 N force is applied?
Answer:
10 mm
Explanation:
We'll begin by calculating the spring constant of the spring. This can be obtained as follow:
Extention (e) = 5 mm
Force (F) = 125 N
Spring constant (K) =?
F = Ke
125 = K × 5
Divide both side by 5
K = 125 / 5
K = 25 N/mm
Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:
Force (F) = 250 N
Spring constant (K) = 25 N/mm
Extention (e) =?
F = Ke
250 = 25 × e
Divide both side by 25
e = 250 / 25
e = 10 mm
Thus, the spring will stretch 10 mm when a 250 N force is applied.
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling at 750 m/s. a) How much work was done by the rocket? b) What is the magnitude of the acceleration of the rocket? c) How long did the flight take?
Answer:
797700000 J
Explanation:
From the question,
The work done by the rocket, is given as,
W = Ek+Ep............. Equation 1
Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.
Ep = mgh............ Equation 2
Ek = 1/2mv²............. equation 3
Substitute equation 2 and equation 3 into equation 1
W = mgh+1/2mv².............. Equation 4
Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.
Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²
Substitute into equation 4
W = 2000(12000)(9.8)+1/2(2000)(750²)
W = 235200000+562500000
W = 797700000 J
A 6 kg box with initial speed 8 m/s slides across the floor and comes to a stop after 2.4 s. A) What is the coefficient of kinetic friction?B) How far does the box move? C) You put a 5 kg block in the box, so the total mass is now 11 kg, and you launch this heavier box with an initial speed of 7 m/s. How long does it take to stop?
Answer:
A. Coefficient of kinetic friction, μ = 0.34
B. The box moves a distance of 9.64 m before coming to a stop
C. The heavier box will stop after 2.1 seconds
Explanation:
a. The coefficient of kinetic or sliding friction is given as: μ = F/R
where applied force, F = m × (∆v)/t
∆v = v - u
∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s
R = normal reaction = m×g
where g = 9.8 m/s²
Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g
μ = ∆v/g×t
μ = 8 / 9.8 × 2.4
μ = 0.34
b. Using the equation v² = u² + 2as to calculate the distance travelled by the box
where v = 0 m/s; u = 8.0 m/s; a = ? s = ?
From F = ma = μR
a = μR/m = (μ × m × g)/m
a = μg
a = 0.34 × 9.8
a = 3.32 m/s²
This is negative acceleration or deceleration
Substituting in the equation of motion
8² + 2 × -3.32 × s = 0
-6.64s = -64
s = 9.64 m
Therefore, the box moves a distance of 9.64 m before coming to a stop
c. The coefficient of friction is independent of mass.
Using the formula in (a): μ = ∆v/g×t
t = ∆v/μg
t = 7/0.34 × 9.8
t = 2.10 s
Therefore, the heavier box will stop after 2.1 seconds
The coefficient of kinetic friction of the box is 0.34.
The distance traveled by the box is 9.6 m.
The time taken for the heavier box to stop is 2.1 s.
The coefficient of kinetic frictionThe coefficient of kinetic friction of the box is calculated as follows;
[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]
The distance traveled by the boxThe distance traveled by the box is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]
The time taken for the heavier box to stop is calculated as;
[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]
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What two air masses creates hurricanes?
Answer:
The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.
Explanation:
Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.
Answer:
air
Explanation:
How much energy is required to move 2 electrons through a potential difference of 1.0 x 10^ 2 volts?
The energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is -3.2 x 10⁻¹⁷ joules.
What is the potential difference?The potential difference, also known as voltage, is the difference in electrical potential energy per unit of charge between two points in an electrical circuit or an electric field.
In simpler terms, the potential difference is the amount of energy required to move a unit of electric charge from one point to another in an electric field or an electrical circuit. It is measured in volts (V) and can be calculated using the equation:
V = W/Q
Where V = is the potential difference,
W = is the work done in moving the charge Q from one point to another,
Q =is the amount of charge that is moved.
Potential difference is an essential concept in electrical engineering and physics, as it governs the flow of electric current in a circuit and determines the behavior of electrical devices such as resistors, capacitors, and batteries.
Here in this question,
The energy required to move an electron through a potential difference is given by the formula:
E = qV
Where
E = is the energy required,
q = is the charge of the electron,
V = is the potential difference.
The charge of one electron is -1.6 x 10⁻¹⁹coulombs.
Therefore, for two electrons, the total charge is:
q = 2 x (-1.6 x 10⁻¹⁹coulombs) = -3.2 x 10⁻¹⁹ coulombs
The potential difference is given as 1.0 x 10² volts.
Now, the energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is:
E = (-3.2 x 10⁻¹⁹ coulombs) x (1.0 x 10² volts) = -3.2 x 10⁻¹⁷ joules.
Note: that the negative sign indicates that the electrons lose potential energy as they move through the potential difference.
Therefore, The required energy is -3.2 x 10⁻¹⁷ joules.
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4 Two friction disks A and B are to be brought into contact withoutslipping when the angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at time t = 0 and is given a constantangular acceleration with a magnitude α. Disk B starts from rest attime t = 2 s and is given a constant clockwise angular acceleration,also with a magnitude α. Determine (a) the required angular acceleration magnitude α, (b) the time at which the contact occurs
This question is incomplete, the missing image is uploaded along this answer below;
Answer:
a) the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) the time at which the contact occur is 8 seconds
Explanation:
Given the data in the question;
first we convert the given angular velocity to rad/s
angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s
so
ωA = 8π rad/s
next we determine angular acceleration at point A; so
ωA = at
8π rad/s = at -------let this be equation
thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.
Next we determine the velocity of point C;
Vc = rA × ωA
where Vc is velocity at point C, rA is radius of A ( 150/1000)m, { from the diagram }
so we substitute
Vc = 0.15m × 8π
Vc = 1.2π m/s
for angular velocity at point B;
Vc = rB × ωB
where rB is the radius of B ( 200/1000)m
we substitute
1.2π = 0.2 × ωB
ωB = 1.2π / 0.2
ωB = 6π rad/s
Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.
Now,
a) Determine the required angular acceleration magnitude α
we find the the angular acceleration of disk B after 2 seconds, using the expression;
ωB = at
where angular acceleration is a and t is time ( t - 2)
we substitute
ωB = at
6π = a( t - 2) -------- let this be equation 2
now, lets substract equation 1 form equation 2
(6π = a( t - 2)) - (8π = at)
(6π = at - 2a) - ( 8π = at)
-2π = 0 + -2a
2π = 2a
a = 2π/2
a = π rad/s² or 3.14 rad/s²
Therefore, the required angular acceleration magnitude α is π rad/s² or 3.14 rad/s²
b) determine the time at which the contact occurs;
from equation 1
8π = at
we substitute in the value of a
8π = π × t
t = 8π / π
t = 8 seconds
Therefore, the time at which the contact occur is 8 seconds
Allison and Heather are going to conduct an experiment to see whether salt affects the growth of plants. They assemble five groups of identical plants and give the plants in each group water with a different salt concentration. What is the outcome variable (dependent variable) for their experiment?
A. Salt concentration in plant tissue
B. Salt concentration in plant water
C. Amount of water absorbed by plants
D. Average mass of plants in each group
Answer:b
Explanation:
Guess
1. With the exception to water, matter (expands, contracts) when it gets
hotter. *
A)Expands
B)Contracts
EXAMPLE
• A patient lying horizontally on a hospital bed is found to have an
enlargement of his blood vessel where the walls have weakened. The
cross-sectional area of the enlargement is 2.0A where A is the cross-
section of the normal aorta. The normal speed of blood through the
person's aorta is 0.40 m/s, and the density of blood is 1,060 kgm-3
Calculate
. a) the speed of blood in the enlargement.
• b) how much higher the pressure is in the enlargement.
Answer:
Explanation:
a ) The volume of blood flowing per second throughout the vessel is constant .
a₁ v₁ = a₂ v₂
a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .
2A x v₁ = A x .40
v₁ = .20 m /s
b )
Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂
Applying Bernoulli's theorem
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
P₁ - P₂ = 1/2 ρ(v₂² - v₁² )
= .5 x 1060 ( .4² - .2² )
= 63.6 Pa .
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
Which describes the length of the power cord when Shanti gets off the train?
cannot be determined
less than 1.2 m
more than 1.2 m
equal to 1.2 m
Answer:
D. equal to 1.2
Explanation:
on edg
The length of the power cord will be equal to 1.2 m.
Describe about the length of power cord? The train is moving at a speed of 90 km /hr. Train was moving but the person in the train can be considered to be at rest. Shanti is the person travelling on the train. Her cord can be used only by her and the cord length of the phone will be 1.2 m.The length can be measured through the distance.The unit of length is meter.As we know the concept of motion and rest, there only the train in motion, shanti was at rest and shanti's power cord were also in the rest. Power cord length will be determined only at the time of manufacturing.If the power cord length to be change then the crimping process.So, the length will not change suddenly.
The length of the power cord when shanti gets off the train is equal to 1.2 m.
The Correct answer is Option D.
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Suppose that you begin on a street corner and walk one block in one minute. How would you express the rate of your motion?
O One block per minute.
O The information is insufficient.
O Fast
O One minute per block.
Answer:
The first choice, one block per minute.
Will mark BRAINLYEST
Answer:
A. shin guards is that answer
Answer:
Shin Guards Brainlyiest please
Explanation:
What is cutoff wavelength?
Answer:
The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).
Explanation:
Amy is in-line skating. Her mass is 50 kg. She is rolling forward (north) on a flat section of road at 10 m/s. The rolling friction acting on the wheels of her skates is 10 N backward (south). Air resistance creates a 15 N force also acting backward (south) on her. If these are the only horizontal forces acting on Amy. What is her horizontal acceleration as a result of these forces?
Answer:
The right solution is "0.50 m/s²". A further explanation is provided below.
Explanation:
The given values are:
Mass,
m = 50 kg
Speed,
v = 10 m/s
Rolling friction acting backward (south),
f = 10 N
Air resistance acting backward (south),
[tex]F_T[/tex] = 15 N
The total force acting will be:
⇒ [tex]F = -f-F[/tex]
On substituting the given values, we get
⇒ [tex]=-10-15[/tex]
⇒ [tex]=-25 \ N[/tex]
Now,
⇒ [tex]a = \frac{F}{m}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{-25}{50}[/tex]
⇒ [tex]=-0.50 \ m/s^2[/tex]
The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.
A fish of 108 N is attached to the end of a dangling spring, stretching the spring by 14.0 cm. What is the mass of a fish that would stretch the spring by 23.0 cm?
Answer:
The mass of the fish is 18.1 kg.
Explanation:
Given;
weight of the fish, F = 108 N
extension of the spring by the given weight, x = 14 cm = 0.14 m
First, determine the elastic constant of the spring by applying Hook's law;
F = kx
where;
k is the spring constant
k = F/x
k = 108 / 0.14
k = 771.43 N/m
When the spring is stretched to 23cm, the mass of the fish is calculated as follows;
[tex]F = mg = Kx\\\\m = \frac{Kx}{g} \\\\m = \frac{771.43\times 0.23}{9.8} \\\\m = 18.1 \ kg[/tex]
Therefore, the mass of the fish is 18.1 kg.
I WILL MAKE THE BRAINLEST
6. The front that moved over Roberto's area the last week of the month was humid. Based on the chart and your knowledge of fronts, what kind of front would this most likely be?
Answer choices
A. cold front
B. warm front
C. occluded front
D. stationary front
Answer:
warm front .,
Explanation:
...........
You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating the acceleration of gravity 608 km above the Earth's surface in units of g. (The mass of the Earth is 5.97 x 1024 kg, and the radius of the Earth is 6380 km.)
Answer:
The answer is "83.1%".
Explanation:
Given:
[tex]\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}[/tex]
Using formula:
[tex]\to g_E = G \frac{M_E}{(R_E +h)^2}[/tex]
[tex]\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\[/tex]
[tex]\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}[/tex]
Calculating the gravity on the Earth’s surface:
[tex]\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}[/tex] [tex]= \frac{8.15}{9.8} \times 100=83.1 \%[/tex]
Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
what is costant error
Answer:
In a scientific experiment, a constant error -- also known as a systematic error -- is a source of error that causes measurements to deviate consistently from their true value.
Explanation:
If a biker rides west for 50 miles from his starting position, then turns and bikes back east for 80 miles. What is his displacement?
Answer:
Displacement = 30 miles due east.
Explanation:
Let the distance due west be A
Let the distance due east be B
Given the following data;
A = 50 miles
B = 80 miles
To find the displacement;
Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.
Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.
Displacement = B - A
Displacement = 80 - 50
Displacement = 30 miles due east.
The half-life of iodine-131 is 13 hours. If a sample of radium-226 has an original
activity of 400 Bg, what will its activity be after:
i) 26 hours?
ii) 39 hours?
iii) 52 hours
Answer:
I. 100 Bg
II. 50 Bg
III. 25 Bg
Explanation:
I. Determination of the activity after 26 hours.
We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t½) = 13 hours
Time (t) = 26 hours
Number of half-lives (n) =?
n = t / t½
n = 26 / 13
n = 2
Finally, we shall determine remaining activity
Original activity (N₀) = 400 Bg
Number of half-lives (n) = 2
Activity remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 100 Bg
II. Determination of the activity after 39 hours.
We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t½) = 13 hours
Time (t) = 39 hours
Number of half-lives (n) =?
n = t / t½
n = 39 / 13
n = 3
Finally, we shall determine remaining activity.
Original activity (N₀) = 400 Bg
Number of half-lives (n) = 3
Activity remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2³ × 400
N = 1/8 × 400
N = 50 Bg
III. Determination of the activity after 52 hours.
We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t½) = 13 hours
Time (t) = 52 hours
Number of half-lives (n) =?
n = t / t½
n = 52 / 13
n = 4
Finally, we shall determine remaining activity
Original activity (N₀) = 400 Bg
Number of half-lives (n) = 4
Activity remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2⁴ × 400
N = 1/16 × 400
N = 25 Bg
