Answer:
Reverse check the answer
Explanation:
I believe it is very important that once someone is done with any calculation, the person ought to go over the calculations again. And even, recheck the answer in inverted form.
This is so because while doing the calculations, we can possibly make errors that we won't notice until after submission. Knowing 2 * 3 = 6, but writing 2 * 3 = 5 in the course of calculations can happen to anybody. So therefore, cross checking and reverse checking is needed
When you sweat, what is the external stimuli? I need help asap.
Answer:
An External Stimulus is a stimulus that comes from outside an organism. Examples: You feel cold so you put on a jacket. When you sweat, the external stimulus is either you're anxious or hot.
Explanation:
hope it helps! <3
Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons is this charge?
Answer:
[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]
Explanation:
Coulomb's Law
The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:
[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]
Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:
[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]
Substituting values:
[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]
[tex]q = 1.05\cdot 10^{-5}~c[/tex]
This charge corresponds to a number of electrons given by the elementary charge of the electron:
[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]
Thus, the charge of any of the spheres is:
[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]
[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]
A person standing on top of a 30.0 m high building throws a ball with an initial velocity of 20. m/s at an angle of 20.0° below horizontal. How far from the base of the building will the ball land?
Answer:
Explanation:
horizontal component of velocity of throw = 20 cos20 = 18.8 m /s
vertical downwards component = 20 sin20 = 6.84 m /s
time to displace by height 30 m = t , initial velocity u = 6.84 m /s
h = ut + 1/2 gt²
30 = 6.84 t + .5 x 9.8 t²
4.9 t² + 6.84 t - 30 = 0
t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9
= - 6.84 ±√( 46.78 + 588 ) / 9.8
= - 6.84 ±√(634.78 ) / 9.8
= - 6.84 ±25.2 / 9.8
= 1.87 s
horizontal displacement in 1.87 s
= 18.8 x 1.87
= 35.15 m .
It has been suggested that rotating cylinders about 12.5 mi long and 6.25 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?
Answer:
w = 0.0436 rad/s
Explanation:
Radius of the cylinder is
3.125 * 1609m/mile = 5028 m
Using the angular acceleration formula, we have
a = v²/r = w²r.
Since we're interested in the angular speed, we make w the subject of the formula, so that
w² = a/r
w = √a/r
w = √(9.8/5028)
W = √0.0019049
w = 0.0436 rad/s
Then we can conclude that the angular speed of the rotation is w = 0.0436 rad/s
A crate is pulled due south with a force of 350. N. What other force must be applied if the
net force on the crate is 425 N due north? Enter the magnitude (with units) and direction
(north, south, east, west).
Answer:
775 N due North.
Explanation:
If the crate is pulled South with 350 N force, and the net force on the crate results into 425 N due North, then the other force (F) acting must be larger than the 350 N, and pointing North:
F - 350 N = 425 N
F = 425 N + 350 N = 775 N due North.
Is a windmill that is not running, potential or kinetic?
A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.120. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.559 m/s, what is her speed at the bottom?
Answer:
a
[tex]H =212.6 \ J[/tex]
b
[tex]v = 7.647 \ m/s[/tex]
Explanation:
From the question we are told that
The child's weight is [tex]W_c = 287 \ N[/tex]
The length of the sliding surface of the playground is [tex]L = 7.20 \ m[/tex]
The coefficient of friction is [tex]\mu = 0.120[/tex]
The angle is [tex]\theta = 31.0 ^o[/tex]
The initial speed is [tex]u = 0.559 \ m/s[/tex]
Generally the normal force acting on the child is mathematically represented as
=> [tex]N = mg * cos \theta[/tex]
Note [tex]m * g = W_c[/tex]
Generally the frictional force between the slide and the child is
[tex]F_f = \mu * mg * cos \theta[/tex]
Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as
[tex]F =m* g sin(\theta) - F_f[/tex]
Here F is the resultant force and it is represented as [tex]F = ma[/tex]
=> [tex]ma = m* g sin(31.0) - \mu * mg * cos (31.0)[/tex]
=> [tex]a = g sin(31.0)- \mu * g * cos (31.0)[/tex]
=> [tex]a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)[/tex]
=>[tex]a = 4.039 \ m/s^2[/tex]
So
[tex]F_f = 0.120 * 287 * cos (31.0)[/tex]
=> [tex]F_f = 29.52 \ N[/tex]
Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as
[tex]H = F_f * L[/tex]
=> [tex]H = 29.52 * 7.2[/tex]
=> [tex]H =212.6 \ J[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
=> [tex]v^2 = 0.559^2 + 2 * 4.039 * 7.2[/tex]
=> [tex]v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}[/tex]
=> [tex]v = 7.647 \ m/s[/tex]
At 20 oC the densities of fresh water and ethyl alcohol are, respectively, 998 and 789 kg/m3. Find the ratio of the adiabatic bulk modulus of fresh water to the adiabatic bulk modulus of ethyl alcohol at 20 oC.
Answer:
The ratio is [tex]\frac{B_1}{B_2} = 1.265[/tex]
Explanation:
From the question we are told that
The density of fresh water is [tex]\rho__{f}} = 998 \ kg/m^3[/tex]
The density of ethanol is [tex]\rho_{e} = 789 \ kg /m^3[/tex]
Generally speed of a wave in a substance is mathematically represented as
[tex]v = \sqrt{\frac{B}{\rho} }[/tex]
Here B is the adiabatic bulk modulus of the substance while [tex]\rho[/tex] is the density of the substance
So at constant wave speed
[tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]
=> [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]
=> [tex]B_1 \rho_2 = B_2\rho_1[/tex]
=> [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]
Here [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex] and [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]
So
=> [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]
=> [tex]\frac{B_1}{B_2} = 1.265[/tex]
A ball is launched from ground level at 20 m/s at an angle of 40° above the
horizontal. A) How long the ball is in the air? B)What is the maximum
height the ball can reach?
(a) The ball's height y at time t is given by
y = (20 m/s) sin(40º) t - 1/2 g t ²
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :
0 = (20 m/s) sin(40º) t - 1/2 g t ²
0 = t ((20 m/s) sin(40º) - 1/2 g t )
t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 g t
t = (40 m/s) sin(40º) / g
t ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So
0² - ((20 m/s) sin(40º))² = 2 (-g) y
where y in this equation refers to the maximum height of the ball. Solve for y :
y = ((20 m/s) sin(40º))² / (2g)
y ≈ 8.4 m
Which one would be felt louder: a sound with a SIL of 60dB and a frequency of 40Hz or a sound with a SIL of 60dB and a frequency of 100Hz ? a) they both will appear to have the same loudness b) sound with a SIL of 60dB c) sound with a SIL of 60dB and a frequency of 1000Hz d) whichever is sounded first e) cannot say and a frequency of 40ã
Answer:
a) they both will appear to have the same loudness
Explanation:
When we talk about how loud a sound is , we describe it in decibels (dB). The value of this unit on measurement in this question is what we are going to build our answer on.
Their sound intensity level is the same at 60db even though each of these separate sounds have different frequencies of 40Hz and 100Hz respectively.
The both of them will therefore have the same loudness actually as their sound intensity level has no difference bat 60db each.
If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the internal energy of the gas?
Answer:
The change in the internal energy of the gas 1,595 J
Explanation:
The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)
This is summarized in the expression:
ΔU= Q - W
where the heat absorbed and the work done by the system on the environment are considered positive.
Taking these considerations into account, in this case:
Q= 500 cal= 2,092 J (being 1 cal=4.184 J) W=500 JReplacing:
ΔU= 2,092 J - 500 J
ΔU= 1,592 J whose closest answer is 1,595 J
The change in the internal energy of the gas 1,595 J
At what speed does a 2000 kg compact car have the same kinetic energy as a 18000 kg truck going 21 km/hr?
Answer:
v = 17.5 m/s = 63 km/h
Explanation:
The general expression for the kinetic energy of one moving object is as follows:[tex]K = \frac{1}{2}*m *v^{2} (1)[/tex]
where m = mass of the object, v= speed of the object.
In order to get the value of the kinetic energy of the truck in Joules, we need to convert km/hr to m/s first, as follows:[tex]21 km/hr * \frac{1 hr}{3600s}*\frac{1000m}{1 km} = 5.83 m/seg (2)[/tex]
Now, replacing (2) and m = 18000 kg in (1), we get:[tex]K = \frac{1}{2}*18000 kg *(5.83m/s)^{2} = 306250 J (3)[/tex]
This value must be the same for the 2000 kg compact car, so we can write:[tex]K = 306250 J = \frac{1}{2}*2000 kg *v^{2} (4)[/tex]
Solving for v, we get:[tex]v = \sqrt{\frac{306250}{1000} (m/s)2} = 17. 5 m/s = 63 km/h (5)[/tex]
Which energy transformation occurs after a skydiver reaches terminal velocity? Gravitational potential energy transforms into thermal energy. Gravitational potential energy transforms into kinetic energy. Kinetic energy transforms into thermal energy. Kinetic energy transforms into gravitational potential energy. The answer is A. just took it
Answer:
a
Explanation:
The energy transformation occurs after a skydiver reaches terminal velocity is follows as;
A. Gravitational potential energy transforms into thermal energy.
B. Gravitational potential energy transforms into kinetic energy.
What is the gravitational potential energy?The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:
U = mgh
where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground.
The skydiver gravitational potential energy decreases as the altitude decreases and his kinetic energy store increases as his speed increases.
When a skydiver jumps out of a plane, the energy transfers take place as;
The skydiver's kinetic energy store increases as their speed increases and the thermal store of the air and the skydiver increases, as there is friction between the skydiver and the air particles.
In the given situation, both options A and B are correct.
Learn more about gravitational potential energy;
https://brainly.com/question/11033151
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A motorcycle moving at a constant velocity suddenly accelerates at a rate of 4.0 m/s^2 to a speed of 35 m/s in 5.0 s. What was the initial speed of the motorcycle?
Answer:
15m/sExplanation:
Step one:
given data
acceleration a=4m/s^2
final velocity v=35m/s
time t=5 seconds
Step two:
Required
initial velocity u=?
Applying the first equation of motion
v=u+at
substituting we have
35=u+4*5
35=u+20
u=35-20
u=15 m/s
The initial velocity is 15m/s
A track star runs a 100m race in 12s what is the velocity of the runner?
Answer:
8.33 m/s
Explanation:
v=d/s, velocity = displacement/ time
Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?
Answer:
The value is [tex]w = 0.1167 \ rev/second[/tex]
Explanation:
From the question we are told that
The rate at which the plate rotates is [tex]w =7.0 \ rev/min[/tex]
Generally the revolution per second is mathematically represented as
[tex]w = \frac{7.0}{60}[/tex]
=> [tex]w = 0.1167 \ rev/second[/tex]
What was used as a basket at each end of the gym?
Group of answer choices
bucket
hula hoop
apple basket
peach basket
Answer: apple basket ✨
Explanation:
why is The sum of two vectors has the smallest magnitude when the angle between these two vectors is 180t
Answer:
C = A - B
Explanation:
The addition of vectors takes into account the magnitude of each vector and its direction, so when adding two vectors, the result depends on the direction of the vectors.
* If the vectors have the same direction the result is maximum
C = A + A
* if the vectors have 90 between them, the magnitude of the result is given by the Pythagorean Theorem
C = √(A² + B²)
* if the vectors have 180º between them the result is minimal
C = A - B
We can also perform this sum graphically, where the resulting vector goes from the origin of the first vector to the tip of the last one, it can clearly be seen that when the vectors are antiparallel (180º angle) the magnitude is minimal
Please help me! Will give brainliest.
Answer:
B. Boats need to float.
An artificial satellite in a circular orbit around the Sun has a period of 8 years. Determine the ratio of the satellite's orbital radius to that of the earth's orbital radius. Assume that the earth's orbit around the Sun is circular.
Answer:
The ratio is [tex]R_c:R_e = 4 : 1[/tex]
Explanation:
From the question we are told that
The period of the satellite is [tex]T_c = 8 \ years[/tex]
Generally the period of earth around the sun is [tex]T_e = 1 \ year[/tex]
Generally from Kepler's third law , which is mathematically represented as
[tex]\frac{T_c ^2}{T_e^2} = \frac{R_c^3}{R_e^3}[/tex]
Here [tex]R_c[/tex] is the radius of the orbit which the satellite rotate around the sun
[tex]R_e[/tex] is the radius of the orbit which the earth rotate around the sun
=> [tex]\frac{R_c^3}{R_e^3} = [\frac{8}{1} ]^2[/tex]
=> [tex]\frac{R_c}{R_e} = \sqrt[3]{[\frac{8}{1} ]^2}[/tex]
=> [tex]\frac{R_c}{R_e} = \frac{4}{1 }[/tex]
=> [tex]R_c:R_e = 4 : 1[/tex]
A 5 kg projectile is shot with an initial velocity of 8 m/s from a height of 10 meters. If the acceleration due to gravity is 9.8 m/s^2, what is the total initial energy of the projectile?
Answer: 650 joules
Explanation:
Total energy (kinetic + potential)
1/2×mv^ + mgh
1/2×5×8×8+ 5×9.8×10
1/2×320 + 5×98
320/2 + 480
320/2 + 980/2 (multipling to make the denominator equal)
1300/2 = 650 joules
Hope it helps, if you don't understand try doing it yourself with the formula.
:)
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2 (which agrees with the first answer option given.
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, the first answer option is the correct answer.
Image formed by a plane mirror is
Answer: A plane mirror always forms a virtual image (behind the mirror). The image and object are the same distance from a flat mirror, the image size is the same as the object size, and the image is upright.
Explanation:
A golfer hits the ball off the tee at an angle of thirty-five degrees from the horizontal with a speed of 46 m/s. It lands on the green, which is elevated 5.50 m higher than the tee. How much time elapsed from when the ball was hit to when it landed on the green?
Answer:
t = 5.16 seconds.
Explanation:
The flight time can be found using the following equation:
[tex] y_{f} = y_{0} +v_{0y}*t - \frac{1}{2}gt^{2} [/tex]
Where:
t: is the time
g: is the gravity = 9.81 m/s²
y₀: is the initial height = 0
[tex]y_{f}[/tex]: is the final height = 5.50 m
[tex]v_{0y}[/tex]: is the initial vertical velocity = v*sin(35)
v: is the speed = 46 m/s
[tex] 5.50 = 46*sin(35)*t - \frac{1}{2}9.81*t^{2} [/tex]
By solving the above cuadratic equation we have:
t₁= 0.22 s and t₂= 5.16 s
We will take the solution equal to 5.16 seconds, since in 0.22 seconds (very short time) the ball is going up and in 5.16 seconds it landed on the green.
Therefore, 5.16 seconds have passed since the ball was hit until it landed.
I hope it helps you!
If a person Travels 100 metre due east and then returns to the same place his total displacement is 200. (needed ASAP)
A. True
B. False
Distance is the total path covered by the object
Here, 200 m is the distance covered by the person and NOT the displacement
Displacement of an object is nothing more than the shortest path between the initial and the final point
If the person travelled 100m and came back, his initial and final point will remain the same which means that he will have a displacement of 0 m
A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative relative to the outer surface. How much work is required to eject a positive sodium ion (Na ) from the interior of the cell?
Answer:
The workdone is [tex]W = 1.712 *10^{-20 } \ J[/tex]
Explanation:
From the question we are told that
The potential difference is [tex]V = 107 mV = 107 *10^{-3} \ V[/tex]
Generally the charge on [tex]Na^{+}[/tex] is [tex]Q_{Na^{+}} = 1.60 *10^{-19 } \ C[/tex]
Generally the workdone is mathematically represented as
[tex]W = Q_{Na^{+}}V[/tex]
=> [tex]W = 1.60 *10^{-19 } * 107 *10^{-3}[/tex]
=> [tex]W = 1.712 *10^{-20 } \ J[/tex]
What is the picture called that shows ALL the forces acting on an object?
Answer:
Free Body Diagram
Explanation:
Such picture is called the "free body diagram"
A rock rolls down a hill. Which form of energy is this an example of? (2 points)
a
Chemical
b
Electrical
c
Mechanical
d
Thermal
Answer:
c.Mechanical
Explanation:
When does acceleration due to gravity equal 9.8 m/s downward?
Select one:
a. always
b. when an object has been dropped downward
c. when an object has been thrown upward
d. at the peak of a thrown object's flight
Answer:
b
Explanation:
i took the quiz i think its right
Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to propel itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water. Part A What is the magnitude of the thrust force on the squid
Answer: see attachment
Explanation: