Answer: 1.2%
Explanation:
Given
If one borrows $12,250
and give $147 interest on it
Then the interest is given from the formula
[tex]\Rightarrow S.I=\dfrac{P\times R\times T}{100}\\\\\Rightarrow 147=\dfrac{12,250\times R\times 1}{100}\\\\\Rightarrow R=\dfrac{147}{122.50}\\\\\Rightarrow R=1.2\%[/tex]
Thus, the annual rate of interest is 1.2%
True or False. A person who is nearsighted cannot see objects that are close to them clearly.
Answer:
false
Explanation:
hope it works
A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .
Required:
How far from the second lens is the final image of an object infinitely far from the first lens?
Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
What is mirror equation?The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
https://brainly.com/question/3229491
Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?
1 airplane
travel due north at 300 km while another airplane travels Due South and 300 km are there speed the same or their velocities the same
Answer:
Explanation:
Speed is scalar and velocity is vector. Vector values imply direction as well as magnitude. Therefore, speed and velocity are not the same. The speeds of these 2 planes are the same at 300km/hr, but the velocity of the plane traveling north is +300km/hr while the velocity of the plane traveling south is -300km/hr if we define north as positive and south as negative.
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1177 g, and its velocity is +0.681 m/s after the bullet passes through it. The mass of the second block is 1626 g.
(a) What is the velocity of the second block after the bullet imbeds itself?
(b) Find the ratio of the total kinetic energy after the collision to that before the collision.
Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=[tex]4.67\times 10^{-3} kg[/tex]
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,[tex]m_1=1177g=1.177kg[/tex]
Mass of block 2,[tex]m_2=1626 g=1.626 kg[/tex]
Velocity of block 1,[tex]v_1=0.681m/s[/tex]
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
[tex]mv=m_1v_1+(m+m_2)v_2[/tex]
[tex]4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2[/tex]
[tex]1.66719=0.801537+1.63067v_2[/tex]
[tex]1.66719-0.801537=1.63067v_2[/tex]
[tex]0.865653=1.63067v_2[/tex]
[tex]v_2=\frac{0.865653}{1.63067}[/tex]
[tex]v_2=0.531m/s[/tex]
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
[tex]K_i=\frac{1}{2}mv^2[/tex]
[tex]k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)[/tex]
[tex]k_i=297.59 J[/tex]
Final kinetic energy after collision
[tex]K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2[/tex]
[tex]K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2[/tex]
[tex]K_f=0.5028 J[/tex]
Now, he ratio of the total kinetic energy after the collision to that before the collision
=[tex]\frac{k_f}{k_i}=\frac{0.5028}{297.59}[/tex]
=0.00169
1 point
Q.29. A stone has a weight of 5.7 N.
The gravitational field strength g is 10
N/kg.What is the mass of the stone?
O A 0.57 kg
Answer:
weight/mass = gravitational field strength
Given :
Weight of stone = 5.7 N
Gravitational field strength (g) = 10 N/kg
Taking Mass of stone x
=> 5.7/x = 10
x = 10 * 5.7
x = 57 kg
Therefore mass of stone is 57 kg
A jogger moves from a position x =
4.0 m to a position of x = 16.0 m in
4.0 s. What was her average velocity?
(Unit = m/s)
Don't forget: velocities and displacements to
the right are +, to the left are -,
Please help me!
Answer:
3 m/s
Explanation:
We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:
Initial displacement (d₁) = 4 m
Final displacement (d₂) = 16 m
Change in displacement (Δd) =?
Δd = d₂ – d₁
Δd = 16 – 4
Δd = 12 m
Finally, we shall determine the determine the average velocity. This can be obtained as follow:
Change in displacement (Δd) = 12 m
Time (t) = 4 s
Velocity (v) =?
v = Δd / t
v = 12 / 4
v = 3 m/s
Thus, the average velocity of the jogger is 3 m/s
7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The
driver of the car applies the brakes after a reaction time of 0.5 s and the car slows down at a rate of
10 ms. What happens?
Answer: Car collide with man
Explanation:
Given
Speed of car is [tex]u=30\ m/s[/tex]
Distance of the man from the car is [tex]s=55\ m[/tex]
Reaction time [tex]t_r=0.5\ s[/tex]
Rate of deceleration [tex]a_d=-10\ m/s^2[/tex]
Distance traveled in the reaction time [tex]d_o=30\times 0.5=15\ m[/tex]
Net effective distance to cover [tex]d=55-15=40\ m[/tex]
Distance required to stop the car
[tex]\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m[/tex]
Require distance is more than that of net effective distance. Hence, car collides with the man.
Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?
Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.
Explanation:
It is known that formula for path difference is as follows.
[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex] ... (1)
where, n = 0, 1, 2, and so on
As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.
[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]
Hence, path difference is as follows.
[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]
For lowest frequency, the value of n = 0.
[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]
[tex]\lambda = 4 \Delta L[/tex]
where,
[tex]\lambda[/tex] = wavelength
The relation between wavelength, speed and frequency is as follows.
[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]
where,
[tex]\nu[/tex] = speed
f = frequency
Substitute the values into above formula as follows.
[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]
Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.
Suppose you walk 13.0 m straight west and then 25.0 m straight south. How far are you from your starting point (in m)
Answer:
28.2 m
Explanation:
Applying,
Pythagoras theorem,
a² = b²+c²............... Equation 1
Where a = The distance from my starting point to my current point, b = distance walked due west, c = distance walked due south
From the question,
Given: b = 13 m, c = 25 m
Substitute these values into equation 1
a² = 13²+25²
a² = 169+625
a² = 794
a = √794
a = 28.18 m
a ≈ 28.2 m
7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration
Answer
NB:
- speed, U is measure in m/s
- acceleration, a is measured in m/s²
-time t in seconds , s
Therefore conversation must be made
Speed U = 20km/hrs
=20km÷1hr
But 20km= 20×1000=20000m
1hr= 1×60min×60sec=3600s
U=20000÷3600=5.56m/s
a=30km/hrs
=30km÷1hr
But 30km=30×1000=30000
1hr=3600s
a=30000÷3600=8.33m/s²
From the equation of motion
S=Ut + ½ at².
Where s= distance
S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²
S = 1777.3m
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brakes.
Answer:
V = 11.83 m/s
Explanation:
Given the following data;
Mass = 2000 kg
Force = 10000N
Distance = 14 m
To find the final velocity of the car;
First of all, we would determine the acceleration of the car;
Acceleration = force/mass
Acceleration = 10000/2000
Acceleration = 5 m/s²
Next, we would use the third equation of motion to find the final velocity;
[tex] V^{2} = U^{2} + 2aS [/tex]
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
V² = 0² + 2*5*14
V² = 0 + 140
V = √140
V = 11.83 m/s
Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m
Answer:
English please
Explanation:
I don't understand the question
What type of lens is this?
Tip: A convex lens is thicker at the centre than at the edges.
When light goes through a concave lens, the light that goes through that lens spreads out more due to the structure of the surface of the lens. A concave lens does not form a focal point at the opposing side of the lens. Instead, it spreads out, doing the exact opposite of what a convex lens does. In the picture, we see a lens where the light goes through and unifies at the opposing side, which is what a convex lens does. So, the lens in that image is a convex lens.
3. What type of lens is this?
Answer: Convex
Wrong Answer: Concave
The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
The magnitude of the average total force acting on the object during this time interval is 20 N.
The given parameters:
Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 sThe magnitude of the average total force acting on the object during this time interval is calculated as follows;
[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]
Learn more about resultant force here: https://brainly.com/question/25239010
I need help with physics question.
(D)
Explanation:
Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is
F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)
= 2.0×10^-11 N
A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a speed of 1.7 m/s. Calculate the speed of the bullet before striking the block. Answer in units of m/s
Answer:
the speed of the bullet before striking the block is 302.3 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 28.3 g = 0.0283 kg
mass of the wooden block, m₂ = 5004 g = 5.004 kg
initial velocity of the block, u₂ = 0
final velocity of the bullet-wood system, v = 1.7 m/s
let the initial velocity of the bullet before striking the block = u₁
Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)
0.0283u₁ = 8.5549
u₁ = 8.5549 / 0.0283
u₁ = 302.3 m/s
Therefore, the speed of the bullet before striking the block is 302.3 m/s.
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
50km
Explanation:
So we know we drove 5km a hour.
We drove like this for 10 hours.
To find the total km, we must multiply the time driven by the speed:
time*speed=distance
Plug in our values:
10*5km=50km
So your answer is 50km.
Hope this helps!
E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V
Answer:
hindi ko maintindihan teh
which team won the champions league in 2020 2021
Answer:
Chelsea F.C
Explanation:
Chelsea F.C
Soccer
Which series represents the flow of thermal energy in a heat engine?
Fuel → air inside piston chamber → air outside piston chamber
Fuel → air inside piston chamber → piston movement
Fuel → piston movement
Fuel → air outside piston chamber
Answer:
B - Fuel → air inside piston chamber → piston movement
Explanation:
A cement block accidentally falls from rest from the ledge of a 81.5-m-high building. When the block is 14.0 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way
Answer:
0.41s
Explanation:
solve for t and y
,........
A car starts from rest. If its acceleration is 1.5 m/s2 in 1.5 seconds, then calculate the
distance travelled by it.
Answer:
1.6875 m
Explanation:
Here, we are given that,
Initial velocity (u) = 0 (as it starts from rest)Acceleration (a) = 1.5 m/s²Time taken (t) = 1.5 sThen,
Distance (s) = ?By using the second equation of motion,
★ s = ut + ½at²
s is distanceu is initial velocityt is timea is acceleration⇒ s = 0(1.5) + ½ × 1.5 × (1.5)²
⇒ s = ½ × 1.5 × 2.25
⇒ s = ½ × 1.5 × 2.25
⇒ s = ½ × 3.375
⇒ s = 1.6875 m
∴ Distance travelled by it is 1.6875 m.
Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.
Answer:
[tex]V_{rms}=5.6*10^3m/s[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=5100K[/tex]
Generally the equation for RMS Speed is mathematically given by
[tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]
Where
[tex]K=Boltzman's constant[/tex]
[tex]K=1.38*10^{-23}[/tex]
And
[tex]M=molecular mass[/tex]
[tex]M=4*1.67*10^{-27}[/tex]
[tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]
[tex]V_{rms}=5.6*10^3m/s[/tex]
A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket
Answer:
The maximum height reached by the rocket is 486.53 m
Explanation:
Given;
initial velocity of the rocket, u = 0
acceleration of the rocket, a= 20 m/s²
duration of the rocket first motion, t = 4 s
The distance traveled by the rocket before its thrust failed
h₁ = ut + ¹/₂at²
h₁ = 0 + ¹/₂ x 20 x 4²
h₁ = 160 m
The second distance moved by the rocket is calculated as follows;
The velocity of the rocket before its thrust failed;
v = u + at
v = 0 + 20 x 4
v = 80 m/s
This becomes the initial velocity for the second stage
At maximum height, the final velocity = 0
[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]
The maximum height reached by the rocket = h₁ + h₂
= 160 + 326.53
= 486.53 m
Please help I need this done
A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction
Answer:
The kinetic energy and potential energy lost to friction is 2,420 J.
Explanation:
Given;
total mass, m = 40 kg
initial velocity of the girl, Vi = 5 m/s
hight of the hill, h = 10 m
length of the hill, L = 100 m
initial kinetic energy of the girl at the top hill:
[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]
initial potential energy of the girl at the top hill:
[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]
Total energy at the top of the hill:
E = 500 J + 3920 J
E = 4,420 J
At the bottom of the hill:
final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s
hight of the hill = 0
final kinetic energy of the girl at the bottom of the hill:
[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]
final potential energy of the girl at the bottom of the hill:
[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]
Based on the principle of conservation of energy;
the sum of the energy at the top hill = sum of the energy at the bottom hill
The energy at the bottom hill is less due to energy lost to friction.
[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]
Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.
Write down the chemical formula and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate
Answer: The chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.
Explanation:
The chemical formula of sodium bicarbonate is [tex]NaHCO_{3}[/tex].
In this compound, there are 1 sodium atom, 1 hydrogen atom, 1 carbon atom and three oxygen atoms present.
Therefore, the ratio of atoms is 1 : 1 : 1 : 3
Thus, we can conclude that the chemical formula is [tex]NaHCO_{3}[/tex] and the ratio in which the four elements combine with each other in the compound Sodium- bicarbonate is 1 : 1 : 1 : 3.
Are you aware of human rights violation happening in the community and explain
Answer:
Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three. As the masses collide they stick together. Mass 1 sticks to 2, then 1 2 sticks to 3, then 1 2 3 sticks to 4. When the combined 1 2 3 mass strikes mass 4, by what percentage does the speed decrease in %
Answer:
The speed decreases 75%.
Explanation:
Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.For the first collission, only mass 1 is moving before it, so we can write the following equation:[tex]p_{i} = p_{f} = m*v_{o} (1)[/tex]
Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:[tex]p_{f1} = 2*m*v_{1} (2)[/tex]
From (1) and (2) we get:v₁ = v₀/2 (3)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:[tex]p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)[/tex]
Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:[tex]p_{2} = 3*m*v_{2} (5)[/tex]
From (4) and (5) we get:v₂ = v₀/3 (6)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:[tex]p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)[/tex]
Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:[tex]p_{3} = 4*m*v_{3} (8)[/tex]
From (7) and (8) we get:v₃ = v₀/4This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.