Answer:
117.83° F
Explanation:
Using Newton's Law of Cooling which can be expressed as:
[tex]\dfrac{dT}{dt}= k(T-T_1)[/tex]
The differential equation can be computed as:
[tex]\dfrac{dT}{dt}= k(T-70)[/tex]
[tex]\dfrac{dT}{(T-70)}= kdt[/tex]
[tex]\int \dfrac{dT}{(T-70)}= \int kdt[/tex]
[tex]In|T-70| = kt +C[/tex]
[tex]T- 70 = e^{kt+C} \\ \\ T = 70+e^{kt+C} \\ \\ T = 70 + C_1e^{kt} --- (1)[/tex]
where;
[tex]C_1 = e^C[/tex]
At the initial condition, T(0)= 350
[tex]350 = 70 C_1^{k*0}[/tex]
[tex]350 -70 = C_1[/tex]
[tex]280 = C_1[/tex]
replacing [tex]C_1[/tex]= 280 into (1)
Hence, the differential equation becomes:
[tex]T(t) = 70 + 280 e^{kt}[/tex]
when;
time (t) = 1 hour
T(1) = 250
Since;
[tex]250 = 70 + 280 e^{k*1}[/tex]
[tex]180 = 280e^k \\ \\ \dfrac{180}{280}= e^k[/tex]
[tex]k = In (\dfrac{180}{280})[/tex]
k = -0.4418
Therefore;
T(t) = 70 + 280e^{(-0.4418)}t
After 4 hours, the temperature is:
T(t) = 70 + 280e^{(-0.4418)}4
T(4) = 117.83° F
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
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
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 uniformly charged disk of radius 35.0 cm carries charge with a density of 7.9 x 10-3 C/m2. Determine the magnitude of the electric field on the axis of the disk at 50.0 cm from the center of the disk. Record the answer to the nearest tenth of a MN/C (mega N/C).
Answer:
E= 80.71 MegaN/C
Explanation:
The radius of the disk is 35.0 cm.
The charge density is 7.9 x 10^-3 C/m^2
The distance from the center of the disk is 50.0 cm
The electric field is given by the expression as shown below:
[tex]E=2 \pi k_{e} \sigma\left[1-\frac{x}{\left(R^{2}+x^{2}\right)^{1 / 2}}\right][/tex]
Plugging the values in the above expression,
E = (2 *3.14 * 7.9 *10^-3 * 9 *10^9) *(1-(0.5/(sqrt(0.5^2+ 0.35^2)))
E= 80.71 MegaN/C
True or False. A person who is nearsighted cannot see objects that are close to them clearly.
Answer:
false
Explanation:
hope it works
A 0.50 kg croquet ball is initially at rest on the grass. When the ball is
struck with a croquet mallet, the average force exerted on the ball is 230
N. If the ball's speed after being struck is 3.2 m/s. (a) Calculate the change
in the momentum and (b) the time during which the momentum was
changed, of the croquet ball ? *
Answer:
(a) 1.6 Ns
(b) 6.95ms
Explanation:
mass of ball, m = 0.5 kg
initial velocity, u = 0
force, F = 230 N
final speed, v = 3.2 m/s
(a) According to the formula of momentum
The change in momentum = m (v - u) = 0.5 (3.2 - 0) = 1.6 Ns
(b) According to the Newton's second law, the force acting on the ball is rate of change of momentum of ball. let the time is t.
[tex]F =\frac{m(v-u)}{t}\\\\230 =\frac{1.6}{t}\\\\t =6.95\times 10^{-3} s = 6.95 ms[/tex]
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
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.
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!
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.”
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?
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.
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
what are proton and neutron ?
Proton is a stable subatomic particle occurring in all atomic nuclei, with a positive electric charge equal in magnitude to that of an electron.
Neutron is a subatomic particle of about the same mass as a proton but without an electric charge, present in all atomic nuclei except those of ordinary hydrogen.
Two circular, concentric and coplanar turns of
radii R1 = 30 cm and R2 = 20 cm are traversed by currents
i1 = 5 A and i2 = 2 A, indicated in the figure. being the constant of
magnetic permeability in vacuum μ0 = 4^.10-7 Tm/A,
characterize the magnetic induction vector originating at center O.
Answer:
(check the pic)
Explanation:
hope it helps
Please help I need this done
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
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
,........
What is the reason why many ships and plane ones traveled in this boundary got disappeared and no debris are found in the bermuda triangle?
Karl Kruszelnicki has insisted the reason why so many ships and planes vanish without trace in the area between Bermuda, Florida, Puerto Rico is nothing to do with aliens or fire-crystals from the lost city of Atlantis.
Instead, the Australian scientist ‘revealed’, the high number of disappearances is explained by nothing more supernatural than plain old human error plus bad weather and the fact that lots of planes and ships enter that area of the Atlantic Ocean in the first place.
Which of the following would produce a star with the longest lifespan
1) a star with no mass
2)a star with low mass
3) a star with medium mass 4) a star with high mass 5)these options would not all be the same
Answer:
1
Explanation:
because stars with the lowest Mass live longer because they burn out slower than a star with high mass
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.
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
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.
An object travels 4 km from A to B towards north and then 3 km from B to C towards west. Find its displacement.
QUËSTIONS:-
An object travels 4 km from A to B towards north and then 3 km from B to C towards west. Find its displacement.
ANSWER:- displacement
[tex] {x}^{2} = {3 }^{2} + {4}^{2} \\ {x}^{2} = 9 + 16 \\ {x}^{2} = 25 \\ x = \sqrt{25} \\ x = 5[/tex]
displacement = 5 km
Explanation:
We nee to use triangle law
As displacement is the shortest path travelled, we take A-C
so, D=√4^2+3^2
D=5km
The weather map below shows air pressure reading given in millibars. Points A,B,C and D are locations of earths surface at which location is surface wind speed the lowest
Answer:
c
Explanation:
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.
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
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]
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
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
