Answer:
unloosen the tight attire parts,open all windows for better air circulation,if he/she does not react place your hands on the chest and press gently three times per interval while giving them air through the mouth CPR then when they react place them to lie horizontal face sideways
What is Triangle ill send a picture
I need help with this please
Which pair of objects would be most strongly attracted to each other?
A. A positively charged particle and a negatively charged particle
B. Two positively charged particles
C. Two negatively charged particles
D. A negatively charged particle and a neutral particle
Answer:
Its A
Explanation:
Just did the quiz
The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.
Pair of objects that are most strongly attracted:When there is a positively charged particle & the negatively charged particle so due to this it should be strongly attracted. Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.
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A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?
Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.
Answer:
9.57m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 90m/s
Distance = 423m
Unknown:
Acceleration of the race car = ?
Solution:
To solve this problem, we should apply one of the appropriate motion equations;
V² = U² + 2as
Where V is the final velocity
U is the initial velocity
a is the acceleration
s is the distance
90² = 0² + 2 x a x 423
8100 = 846a
a = 9.57m/s²
in the figure shown if angle i increases slightly angle r will
Answer:
we need the image to do so.
Explanation:
sorry
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds s
Answer:
Explanation:
The question is incomplete.
The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.
s= sin2(pi)t
Acceleration = d²S/dt²
dS/dt = 2πcos2πt
d²S/dt² = -4π²sin2πt
A(t) = -4π²sin2πt
Next is to find acceleration after 4.5 seconds
A(4.5) = -4π²sin2π(4.5)
A(4.5) = -4π²sin9π
A(4.5) = -4π²sin1620
A(4.5) = -4π²(0)
A(4.5) = 0m/s²
Suppose a signal from a vibrating motor is to be sampled discretely with a digital data acquisition system to determine the RPM of the motor. It is known that the maximum possible RPM is 1800. What is the minimum sampling frequency (in Hz) with which the signal can be sampled in order to measure the RPM of the motor
Answer:
f > 60 Hz
Explanation:
According to Nyquist's Sampling Theorem, to be fully reconstructed without any aliasing, the signal must be sampled at least more than twice during the period of the maximum frequency of the signal.In this case, the signal to be sampled has only one frequency.However, as we have the information in RPM, we need to convert this to cycles/sec (Hz) first, as follows:[tex]f = \frac{1800 rev}{min} * \frac{1 min}{60 sec} = 30 Hz[/tex]
Per Nyquist, fs > 2*30 Hz⇒ fs > 60 Hz
What is the force of a 12 kg object that is accelerating 6 m/s
We are given:
Mass of object (m) = 12 kg
acceleration (a) = 6 m/s²
Solving for the Force:
From newton's second law of motion:
F = ma
replacing the variables
F = 12*6
F = 72N
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
A tennis ball is dropped from a roof 16 meters from the ground. How long does it take
for the ball to reach the ground?
3.3 seconds would be the right answer
The time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.
What is Motion?Motion is simply the change in position of an object or particle over time.
From the Second Equation of Motion;
s = ut + (1/2)gt²
Where s is the distance from ground level, u is initial velocity, t is time elapsed and g is acceleration due to gravity ( g = 9.8m/s² ).
Given the data in the question;
Since the ball was initially at rest before it was dropped.
Initial velocity u = 0Height or distance from gound level s = 16mTime taken to reach the gound t = ?We substitute our values into the expression above.
s = ut + (1/2)gt²
16m = ( 0 × t ) + ( (1/2) × 9.8m/s² × t² )
16m = 0.5 × 9.8m/s² × t²
16m = 4.9m/s² × t²
t² = 16m / 4.9m/s²
t² = 3.2653s²
t = √(3.2653s²)
t = 1.8s
Therefore the time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.
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#SPJ2
How long would it take you to walk 3,962 km from New York to Los
Angeles?
Answer:
913 hours ur welcome :)
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
Define reflection of sound?
In each of the given pairs, choose which element will have the bigger atom. Give reasons for your choice. (a) Mg (atomic number 12) or Cl (atomic number 17) (b) Na (atomic number 11) or K (atomic number 19)
Answer:
Mg (atomic number 12)
K (atomic number 19)
Explanation:
The size of an atom is estimated in terms of its atomic radius.
The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.
Across a period, atomic radii decrease progressively from left to right. This is due to the progressive increase in the nuclear charge without increase in the number of electronic shells. Down a group, atomic radii increase progressively due to the successive shells of electrons being added which have been compensated for by the increase in nuclear charge.Cl is further right of Mg in the third period
K is below Na in the first group
How long is a day in Neptune
Answer: the long day in neptune would be .18383562 years!
Explanation:also for every day is 16 hours
1. Which statement is true of culture?
O Culture do not change or evolve
O Culture refers to the degree in which resources are used to address for problems.
O Culture has little affect on a persons life
Food can reflect culture
A light wave in space is described by the general function: u(r, t) = 1 (2π) 3/2 Z dk A(k) e i(k·r−ωt) . (1) (a) Find the particular solution knowing that at t = 0 the wave is given by
Answer:
hello your question is incomplete attached below is the complete question
Answer : attached below
Explanation:
A) finding the particular solution at t = 0
attached below is the detailed solution of the particular solution knowing that t = 0
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
a box with a Constance velocity has a 5 N of force applied to it from all sides and direction. what will happen to the motion of the box as result?
A-the object will come to rest
b-the velocity of the object will remain the same
c- the velocity of the object will decrease
d- the velocity of the object will increase
Answer:
b-the velocity of the object will remain the same
Explanation:
Forces from opposite sides cancel each other, so there is no net force on the box that would affect its motion. The velocity of the box will remain unchanged.
__
(The box may be crushed, but it will continue in the same direction at the same speed.)
Answer:
b the velocity of the object will remain the same
Explanation:
use your brain:)
The car alarm on a stationary car emits sound waves with a frequency of 450 Hz. If you are moving away from the stationary car at 20 m/s, what alarm frequency do you perceive
Answer:
Explanation:
The frequency of wave is directly proportional to velocity
f = kV
k = f/V
f1/V1 = f2/V2
Given
f1 = 450Hz
V1 = 343m/s
f2 = ?
V2 = 20m/s
Substitute into the formula
450/343 = f2/20
Cross multiply
343f2 = 450×20
343f2 = 9000
f2 = 9000/343
f2 = 26.24Hz
How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______ minutes c) How long does it take a Love wave to travel 7,000 km? ______ minutes d) How long does it take a Rayleigh wave to travel 7,000 km? ______minutes
Answer:
A. 8.64 secs.
B. 14.58 secs.
C. 26.002 secs.
D. 33.46secs.
Explanation:
A. P wave would travel 7000km
p-wave travels on a speed of 13.5km/s
= 7000km/13.5km/s
= 8.64 secs.
B. S-wave time to travel 7000km
s-wave travels on a speed of 8km/s
= 7000km/8km/s
= 14.58 secs.
C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).
= 7000km to miles
= 4349.598m/2.788m/s
= 26.002 secs.
D. Rayleigh wave to travel 7,000 km
10,000m/s ( 2.1667 m/s ).
= 7000km to miles
= 4349.598m/2.1667m/s
= 33.46secs.
A person drops a marble from the top of a skyscraper. After falling four floors the marble has gained a certain speed. How many more floors will the marble have to fall to triple this speed?
a. 8
b. 12
c. 32
d. 48
Answer:
B. 12
Explanation:
4 x 3 = 12
The feeling of weightlessness occurs because _____________________.
there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.
Answer:
there is only a small amount of gravity present.
Explanation:
this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.
Peter is running laps around a circular track with a diameter of 100 meters. If it takes Peter 12 minutes to run 4 laps, how quickly is he running (in meters per second)?
Answer:
v = 1.74 m/s
Explanation:
Given that,
Diameter of a circular track, d = 100 m
Distance covered for the 4 laps,
[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]
Time, t = 12 minutes = 720 s
We need to find the velocity of the peter. It can be calculated as follows :
[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]
So, the speed is running with a velocity of 1.74 m/s.
Peter is running at 1.7453 m/sec.
Given to us,
Diameter of the circular track, D = 100 meters,
Number of laps Peter run, L = 4 laps,
Time taken by Peter, t = 12 minutes,
1 lap = circumference of the circle,
4 laps = 4 x circumference of the circle,
As we know, the circumference of a circle is given by πD.
So, 4 laps = 4 x circumference of the circle,
[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]
Also, we know that 1 minute has 60 sec.
so, 4 minutes = (4 x 60) seconds
Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]
Thus,
[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]
Hence, Peter is running at 1.7453 m/sec.
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how do you Convert 50 g to kg in an equation for physics
Answer:
Divide by 1000
Explanation:
You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?
Answer:
[tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Explanation:
Given that,
The radius of a sphere is (6.45 ± 0.30) cm
Mass of the sphere is (1.79 ± 0.08) kg
Density = mass/volume
For sphere,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]
We can find the uncertainty in volume as follows :
[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]
Uncertainty in mass,
[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]
Now, the uncertainty in density of sphere is given by :
[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]
Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?
A) is the mechanical conserved between a and b? explain
B) is the mechanical energy conserved between b and c ?explain
Answer:
Yes at A the mechanical energy is conserved.
Yes at B the part of mechanical energy is conserved potential energy and kinetic energy and some is lost as frictional force.
Explanation:
Ratio = Eb/ Ea= 1058.3 J/2940 J= 0.3599
Ratio = Ec/ Eb= 0J/ 1058.3 J= 0
At point A the skater is at rest or it is the starting point and the whole energy is due to the position of the skater i.e= mgh = 50 *9.8*6= 2940 J
Since there's no movement there is no Kinetic energy = 0 J
Yes at A the mechanical energy is conserved.
At point B the skater has traveled for some of the distance . It has potential energy and kinetic energy.
Yes at B the part of mechanical energy is conserved as potential energy and kinetic energy.
The total Mechanical energy = 1058.3 J
At point B Total Mechanical energy = PE+ KE
1058.3J = 980 J + 78.3 J
1058.3 J = mgh + 1/2mv²
= 50*2*9.8 + 1/2 *50*(8.85)²
= 980 J + 78.3 J
As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .
2940 J-1058.3 J= 1881.7
At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME all are zero.
(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.
(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.
(c) mechanical energy is conserved between a and b.
(d) mechanical energy is not conserved between b and c.
The given parameters;
mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]The ratio of the mechanical energy at B and mechanical energy at A;
[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]
The ratio of the mechanical energy at C and mechanical energy at A;
[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]
Thus, we can conclude that mechanical energy is conserved between a and b.
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]
Thus, we can conclude that mechanical energy is not conserved between b and c.
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