Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car [tex]V_{c}[/tex] = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
[tex]V_{f}[/tex] = 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
[tex]V_{f}[/tex] = [tex]V_{i}[/tex] + at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
[tex]d_{c}[/tex] = [tex]V_{c}[/tex] × t₁
we substitute
[tex]d_{c}[/tex] = 22.352 × 7
[tex]d_{c}[/tex] = 156.46 m
now distance between polive car and speeding car
Δd = [tex]d_{c}[/tex] - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( [tex]V_{f}[/tex] - [tex]V_{c}[/tex] )
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = [tex]V_{f}[/tex] × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
What is an
example of a force?
Pressure
Power
Energy
Weight
Answer:
weight
Explanation:
weight is the pull of gravity so its a force !!
According to Coulomb's Law, if the distance between two charged particles is doubled, the electric force will be _________. *
Answer: reduced by 1/4
Explanation:
The force will be reduced by 1/4. Try plugging in 2r, then squaring it. You will get 4r^2, which is essentially dividing the force by 4
what's an equation that represents the relationship between speed distance and time
Answer:
The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).
What is the frequency of highly energetic ul-
traviolet radiation that has a wavelength of
124 nm?
The speed of light is 3 x 108 m/s.
Answer in units of Hz.
Frequency = (speed) / (wavelength)
Frequency = (3 x 10⁸ m/s) / (124 x 10⁻⁹ m)
Frequency = 2.42 x 10¹⁵ Hz
A certain sprinter has a top speed of 11.3 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 12.8 m. He is then able to maintain his top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach
h his top speed. What must this distance be if he is to achieve a time of 9.75 s for the race?
Answer:
Explanation:
initial velocity u = 0
final velocity v = 11.3 m /s
distance covered s = 12.8 m
v² = u² + 2 a s
11.3² = 0 + 2 x a x 12.8
a = 4.99 m /s²
again ,
v = u + a t
11.3 = 0 + 4.99 t
t = 2.26 s .
Rest of the sprint will be covered with uniform velocity .
Distance covered = 100 - 12.8 = 87.2 m
speed = 11.3 m /s
time taken = 87.2 / 11.3 = 7.7 s
Total time of 100 m sprint = 7.7 + 2.26 = 9.96 m .
b )
Let the time taken to reach the top speed be t .
acceleration a = 11.3 / t
distance covered s = 1/2 a t²
= .5 x (11.3 / t) x t²
= 5.65 t
Rest of the distance = 100 - 5.65 t
time taken to cover rest of the distance = (100 - 5.65 t ) / 11.3
Total time = (100 - 5.65 t / 11.3 ) + t = 9.75
100 - 5.65 t + 11.3 t = 11.3 x 9.75
100 + 5.65 t = 110.175
5.65 t = 10.175
t = 1.8
acceleration a = 11.3 / t
= 11.3 / 1.8
= 6.278 m /s²
distance covered in 1.8 s
s = 1/2 a t²
= .5 x 6.278 x 1.8²
= 10.17 m .
a body is projected at an angle of 30 degrees to the horizontal with a velocity of 15m/s, calculate the time it takes to reach the greatest height take g=10ms2 and neglect air resistance
Explanation:
The vertical component the velocity of the projectile is 15 m/s x sin 30 = 7.5 m/s.
The body is accelarating downwards at 10 m/s^2.
This means that every second its upward velocity reduces by 10 m/s.
So if the body is travelling upwards at 7.5 m/s then how long does it take for the velocity to become 0?
(7.5 m/s) / (10 m/s^2) = 0.75 s
A north magnetic pole is facing another north magnetic pole with a distance xx. If the distance between the poles becomes 12x12x, what happens to the magnitude of the field energy between them
Answer:
Explanation:
When two north magnetic poles are placed close to each other , they will repel each other . If distance between them is increased due to mutual repulsion from x to 12 x , work is done by the magnetic field . This results in decrease of magnetic field energy .
Hence , when the distance between the poles becomes 12x from x , the magnitude of the field energy between them decreases .
The amount of kinetic energy an object has depends on its mass and its speed.
a. True
b. False
Answer:
True
Explanation:
The kinetic energy of an object is the energy that it possesses due to its movement or motion.
True, the amount of kinetic energy an object has depends on its mass and its speed.
The kinetic energy of a moving object is directly proportional to the square of its velocity and directly proportional to its mass.
ASAP please thank you !
Find O and P
Explanation:
O + 6V = 9V, so O = 3V
P = 9V as it is parallel to the 9V power supply.
What are 3 things you could you do this week to help you connect better with kids in
your classes?
Answer:
In my physics class, something that helps connect better with kids is keeping connected with them always make sure to ask if they understand what you're teaching if they are following because sometimes most kids are to afraid to admit that they are lost, another way to connect with kids is maybe posting surveys to be able to check in with each student especially during this hard times :)
Explanation:
1. start with fun activities.
2. Encourage single-tasking.
3. Designate a learning playing field.
Which of the following best describes what were wrong with the scientists study
Connective Tissue in a tendon is
Match to the correct answers
A:You measure power in
B:Coal is a power source that is
C:Solar is a power source that is
D:You measure work done in
E:Fuels that took millions of years to form are
Renewable
watts
non-renewable
fossils fuels
joules
plsss help me
A plastic rod 1.6 m long is rubbed all over with wool, and acquires a charge of -9e-08 coulombs. We choose the center of the rod to be the origin of our coordinate system, with the x-axis extending to the right, the y-axis extending up, and the z-axis out of the page. In order to calculate the electric field at location A = < 0.7, 0, 0 > m, we divide the rod into 8 pieces, and approximate each piece as a point charge located at the center of the piece.
Solution :
Length of the plastic rod , L = 1.6 m
Total charge on the plastic rod , Q = [tex]$-9 \times 10^{-8}$[/tex] C
The rod is divided into 8 pieces.
a). The length of the 8 pieces is , [tex]$l=\frac{L}{8}$[/tex]
[tex]$=\frac{1.6}{8}$[/tex]
= 0.2 m
b). Location of the center of the piece number 5 is given as : 0 m, -0.09375 m, 0 m.
c). The charge q on the piece number 5 is given as
[tex]$q=\frac{Q}{L}\times l$[/tex]
[tex]$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$[/tex]
= [tex]$-1.125 \times 10^{-8}$[/tex] C
d). WE approximate that piece 5 as a point charge and we need to find out the field at point A(0.7 m, 0, 0) only due to the charge.
We know, the Coulombs force constant, k = [tex]$8.99 \times 10^9 \ N.m^2/C^2$[/tex]
So the X component of the electric field at the point A is given as
[tex]$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$[/tex]
= -126.15 N/C
The Y component of the electric field at the point A is
[tex]$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$[/tex]
= -16.93 N/C
Now since the rod and the point A is in the x - y plane, the z component of the field at point A due to the piece 5 will be zero.
∴ [tex]$E_z=0$[/tex]
Thus, [tex]$E= <-126.15,-16.93,0>$[/tex]
Using your knowledge on personal care products, how does sunscreen
lotion protect your skin from the damaging effect of ultraviolet rays?
Explain why of x-rays and gamma rays are commonly used in
radiotherapy.
Answer:
Ultraviolet rays from sun are very harmful from skin and can cause sunburn and skin diseases especially ultraviolet B rays. A sunscreen lotion act as a protection barrier on the skin that restrict the direct contact of UV rays with skin and filter the harmful rays to enter the skin.
Radiotherapy is a medical therapy use to treat cancer. Radiotherapy commonly uses x-rays and gamma rays because they are high-energy particles or waves that kills or destroys the cancer cells.
Sunscreen lotion is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
The sun reaches us from outer space brings ultraviolet rays to us. Ultraviolet rays are known to have some damaging effects on the skin. One way to protect our skin from this damaging ultraviolet rays is to use sunscreen lotion which is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
X-rays and gamma rays are used in radiotherapy because they are light energy rays which are able to penetrate and destroy malignant cells in the body.
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After a projectile is fired into the air, what is the magnitude of the acceleration
in the y-direction? (Assume no air resistance.)
O A. 9.8 m/s2
O B. 4.9 m/s2
O C. 19.6 m/s2
O D. O m/s2
SUBMIT
Answer: Option A; 9.8 m/s^2
Explanation:
When an object is in the air, and there is no air resistance acting on the object, the only force that will act on the object is the gravitational force (on the vertical axis).
Then, if the only force acting on the object is the gravitational force, the acceleration of the object will be equal to the gravitational acceleration.
We know that the gravitational acceleration is equal to:
g = 9.8m/s^2
Then the acceleration on the vertical axis will be equal to:
a(t) = 9.8m/s^2
The correct option is the first one:
A. 9.8 m/s^2
Which object exerts the action force?
Which object exerts the reaction force?
In what direction does the action force push?
In what direction does the reaction force push?
For answering this question,let us assume that a person is pushing against the walls,so now:
Which object exerts the action force?
PersonWhich object exerts the reaction force?
WallIn what direction does the action force push?
BackwardIn what direction does the reaction force push?
ForwardThe answer varies from different scenarios.
Answer:
diver, diving board, down, and up.
Explanation:
According to the law of conservation of energy, how will the sum of the kinetic
and potential energies compare at the four points?
Can you also answer questions 1-5
Answer:
I think that this ans may help you
find the moment of inertia of a point mass 0.005g at aperpendicular distance of 3m from its axis of rotation.
Answer:
the moment of inertia is 4.5 × 10⁻⁵ kg.m²
Explanation:
Given that;
point mass m = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
perpendicular distance r = 3m
We know that a point mass doesn't have a moment of inertia around its own axis but, but using the parallel axis theorem, a moment of inertia around a distant axis of rotation can be determined using;
[tex]I_{}[/tex] = mr²
so we substitute
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × (3 m)²
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × 9 m²
[tex]I_{}[/tex] = 4.5 × 10⁻⁵ kg.m²
Therefore; the moment of inertia is 4.5 × 10⁻⁵ kg.m²
The moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm² at a perpendicular distance of 3 m.
The moment of inertia of given point mass can be determined by,
[tex]I = mr^2[/tex]
Where,
[tex]I[/tex]- moment of inertia
[tex]m[/tex]- mass = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
[tex]r[/tex] - perpendicular distance = 3 m
Put the values in the formula,
[tex]I = (5 \times 10^{-6}{\rm \ kg}) \times (3 {\rm \ m})^2\\\\I = 5 \times 10^{-6}{\rm \ kg} \times 9 {\rm \ m}\\\\I = 4.5 \times 10^{-5} kgm^2[/tex]
Therefore; the moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm².
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REAL ANSWER OR REPORTED, REAL GETS BRAINLIST
2. What is the weight of:
A 32 kg child
An 82 kg linebacker
A 1000 kg car
3. What is the mass of:
A student who weighs 524 Newtons
A table that weighs 441 Newtons
A hammer that weighs 51 Newtons
4. A 15 kg box sits motionless on the floor.
Draw a force diagram for the box.
Write a vertical Fnet equation for the box.
Determine the weight of the box.
What is the magnitude of the normal force?
Answer:
32kg = 313.6N
82kg = 803.6N
1000kg = 9800N
524N = 53.47kg
441N = 45kg
51N = 5.20kg
for the box diagram the weight of the 15kg box is 147N
the net force will be equal to zero because the box is motionless
Fg goes downward arrow of box 147N
Fnorm goes upward arrow of box 147N
Choose the best explanation from among the following:_________.
1. Charge is conserved, and therefore the mass of the object will remain the same.
2. A positive charge increases an object's mass; a negative charge decreases its mass.
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Answer: 3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Explanation:
Suppose we have an object and we negatively charge it.
Then we are "adding" N electrons to the object.
Remember that the mass of an electron is:
m = 9.11*10^(-31) kg
Then if we add N electrons to an object of mass M, the new mass of the object will be:
Mass = M + N*9.11*10^(-31) kg
So we will have an (almost negligible) increase of the mass of the object.
(Something similar can happen if the object is positively charged, where we remove electrons, then the mass of the object decreases)
Then the correct option is:
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
A car completes a journey in 2 hours. Its average speed over this journey was 45 mph. Calculate the distance travelled by the car in miles.
Marquette King, formerly of the Denver Broncos, is practicingkicking off using a kicking holder with the ball on the ground.For one of the kicks the ball reaches a height of 90.6 m andlands on the ground 53 yds (48.5 m) away. Find the magnitudeof the initial velocity given by his kick to the ball. Treat airresistance as negligible. Hint: Even though the horizontal andvertical motions are independent, there is a quantity that iscommon to both of them.
Answer:
Explanation:
Maximum height reached = 90.6 m . Range = 48.5 m
. Let u be the initial velocity at angle α .
Horizontal range is covered by horizontal component of u .
Vertical height is achieved by vertical component
v² = u² sin² α - 2gh , t is time taken to attain maximum height .
0 = u² sin² α - 2 x 90.6 x 9.8
u² sin² α = 2 x 887.88 -------( 1 )
Range R = u² sin2α / g
48.5 = 2 u² sinα . cos α / 9.8
u² sinα . cos α = 237.65 ----------------------------- ( 2 )
( 1 ) / ( 2 )
Tan α = 2 x 887.88 / 237.65 = 7.47
α = 82⁰
u² sin² α = 2 x 887.88
u² sin² 82 = 2 x 887.88
u² = 1812
u = 42.56 m /s
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time, when they're not sleeping or eating, joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 0.02130.0213 kg, 0.01650.0165 kg, 0.01850.0185 kg, and 0.01930.0193 kg. At this very instant, the x‑x‑ and y‑y‑ components of the mice's velocities are, respectively, (0.675 m/s,−0.417 m/s)(0.675 m/s,−0.417 m/s) , (−0.249 m/s,−0.809 m/s)(−0.249 m/s,−0.809 m/s) , (0.395 m/s,0.953 m/s)(0.395 m/s,0.953 m/s) , and (−0.207 m/s,0.227 m/s)(−0.207 m/s,0.227 m/s) . Calculate the x‑x‑ and y‑y‑ components of Bryce's mice's total momentum, pxpx and pypy .
Answer: [tex]p_{x}=[/tex] 0.0135814 kg.m/s
[tex]p_{y}=-0.000219[/tex] kg.m/s
Explanation: When an object with mass is in motion, we say the object has momentum (p). Momentum is dependent on mass and velocity:
p = m.v
The total momentum of Bryce's mice is calculated as
x-axis
[tex]p_{x}=\Sigma m.v_{x}[/tex]
[tex]p_{x}=[(0.0213)(0.675)+(0.0165)(-0.249)+(0.0185)(0.395)+(0.0193)(-0.207)][/tex]
[tex]p_{x}=[/tex] 0.0135814
At the x-axis, total momentum of Bryce's mice is 0.0135814 kg.m/s.
y-axis
[tex]p_{y}=\Sigma m.v_{y}[/tex]
[tex]p_{y}=[(0.0213)(-0.417)+(0.0165)(-0.809)+(0.0185)(0.953)+(0.0193)(0.227)][/tex]
[tex]p_{y}=-0.000219[/tex]
At the y-axis, total momentum of Bryce's mice is -0.000219 kg.m/s.
1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam can be built to collect water and release it from an elevation difference of 50 m to generate power. Determine how much power can be generated from this river water after the dam is filled
Answer:
the power that can be generated by the river is 117.6 MW
Explanation:
Given that;
Volume flow rate of river v = 240 m³/s
Height above the lake surface a h = 50 m
Amount of power can be generated from this river water after the dam is filled = ?
Now the collected water in the dam contains potential energy which is used for the power generation,
hence, total mechanical energy is due to potential energy alone.
[tex]E_{mech}[/tex] = m(gh)
first we determine the mass flow rate of the fluid m
m = p×v
where p is density ( 1000 kg/m³
so we substitute
m = 1000kg/m³ × 240 m³/s
m = 240000 kg/s
so we plug in our values into ( [tex]E_{mech}[/tex] = m(gh) kJ/kg )
[tex]E_{mech}[/tex] = 240000 × 9.8 × 50
[tex]E_{mech}[/tex] = 117600000 W
[tex]E_{mech}[/tex] = 117.6 MW
Therefore, the power that can be generated by the river is 117.6 MW
Imagine that I have a ping-pong ball and a bowling ball resting on the floor of our classroom. I go up to the bowling ball and give it a push so that it starts rolling. If I give a push of the exact same strength to the ping-pong ball, how will the resulting motion of the ping-pong ball be different
Answer:
the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
Explanation:
We can analyze this exercise from several points of view, if we use Newton's second law
Bowling ball
F = M a₁
pingpongg ball
F = m a₂
as the forces the same
M a₁ = m a₂
a₂ = [tex]\frac{M}{m}[/tex] a₁
Since the mass of the bowling ball is much greater than the ping pong ball,
a₂ »a₁
so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.
If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest
Bowling ball
I = F t = Δp
I = M (v₁ - v₀)
Ping pong ball
I = F t = Δp
I = m (v₂ -v₀)
the impulse itself
M v₁ = m v₂
v₂ = [tex]\frac{M }{ m}[/tex] v₁
so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.
In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
Mary spots Bill approaching the dorm at a constant rate of 2.00 m/s on the walkway that passes directly beneath Mary's window, 17.0 m above the ground. When Bill is L = 130 m away from the point below Mary's window she decides to drop an apple down to him. Ignore the effects of air resistance in your calculations.
Required:
How long should Mary wait to drop the apple if Bill is to catch it 1.65 m above the ground?
Answer:
Mary will have to wait for 63.2 seconds
Explanation:
Time required for the apple to drop from a height of 17.0 m above the ground to 1.65 m above the ground is given by the formula below:
t = √2h/g where h is height through which the object falls, g is acceleration due to gravity
h = 17.0 - 1.65 = 15.35 m
g = 9.8 m/s²
t = √(2 * 15.35/9.8)
t = 1.77 s or approximately 1.8 s
Time taken for bill to get to the point below Mary's window is given below;
time taken = distance/velocity
distance = 130 m; velocity = 2.0 m/s
time taken by Bill = 130/2.0 = 65 s
Therefore, Mary will have to wait for (65 - 1.8) s = 63.2 seconds
Mary should have to wait for 63.2 seconds
Calculation:t = √2h/g
here h is height via which the object falls,
g represent acceleration due to gravity
So,
h = 17.0 - 1.65
= 15.35 m
And,
g = 9.8 m/s²
Now
t = √(2 * 15.35/9.8)
t = 1.77 s or approximately 1.8 s
Now the time taken should be
= 130/2.0
= 65 s
So,
= (65 - 1.8) s
= 63.2 seconds
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A negatively-charged rod is brought close to (but does not touch) two neutral spheres that are in contact with each other but insulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres
Answer:
One sphere obtains a positive charge, while the other obtains a negative charge.
Explanation:
When the negatively-charged rod is brought close to the two neutral spheres in contact with each other, the electrons in both spheres are repelled away from the rod and thus those electrons move into the second sphere farther away from the rod.
The first sphere is now short of electrons.
When the spheres are separated, the first sphere which is short of electrons now carries a net positive charge, while the other sphere which has excess electrons now carries a net negative charge.
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log floating by. Always with the same engine throttle setting, the boater continues to travel upstream for another 1.45 km, which takes her 69.1 min. She then turns the boat around and returns downstream to her starting point, which she reaches at the same time as the same log does. How much time does the boater spend traveling back downstream
Answer:
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex], v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
[tex]t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }[/tex]
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’= [tex]\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}[/tex]
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex]
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’= [tex]\frac{ 1450}{ 0.6499 + 2 \ 1}[/tex]
t ’= 547.19 s
Why is cloning done before an investigation begins??
Answer:Purpose of Cloning
A forensic image of a hard drive captures everything on the hard drive, from the physical beginning to the physical end. ... Hard drives are susceptible to failure. Having two clones gives an investigator one to examine and one to fall back on.Therapeutic cloning involves creating a cloned embryo for the sole purpose of producing embryonic stem cells with the same DNA as the donor cell. These stem cells can be used in experiments aimed at understanding disease and developing new treatments for disease.
Explanation:
hope this helps have a nice night lo❤️❤️❤️l
Cloning
Cloning is defined as a technique used by scientists to make exact genetic copies of living things. Genes, cells, tissues, and even whole animals can all be cloned.
Purpose of CloningEverything on the hard drive is captured by the forensic image. Hard drives can be a failure. If possible, the original drive should be preserved in a safe place and only brought out to reimage if needed. Ideally, all examinations are done on a clone as opposed to the original.
One of the cloning types is Therapeutic cloning which involves creating a cloned embryo for the sole purpose of producing embryonic stem cells with the same DNA as the donor cell. These stem cells can be used in experiments aimed at understanding disease and developing new treatments for disease.
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