Explanation:
It is known that charges tend to move from a higher potential towards lower potential. As a result, charges tend to move from a positive terminal to negative terminal of the battery.
For example, at a certain height water is stored in a water tank. This height is correspondent to the constant potential and water contained is the charge present in the battery.
When a pipe is connected from the tank to the ground then water will start moving towards the ground and similarly if a conducting wire is connected across the battery then charge will also begin to flow in the same direction which is called current.
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?
a. All of the mechanical energy is converted into other forms
b. Some of the mechanical energy is converted into other forms
c. No mechanical energy is converted into other forms
In a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
In a perfect elastic collision, both momentum and kinetic energy of the particles are conserved.
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
[tex]\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2 ^2= \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2[/tex]
When the collision is not perfectly elastic, only momentum is conserved but the kinetic energy is not conserved.
Thus, we can conclude that in a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
Learn more here:https://brainly.com/question/18432099
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
Its Acceleration during the upward Journey ?
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s
A car accelerates from 0-30m/s in 5 s. It has a mass of 1200kg. What force does the engine produce? *
A) 36000N
B) 7200N
C) 60000N
D) 2000N
include explanation please
Answer:
a = ?
u = 0
v = 30
by using v = u + at equation we can find " a "
30 = 0 + 5a
6 m/s = a
by using f = ma equation we can find force produce by engine ,
f = ?
a = 6
m = 1200
f = 1200 × 6
f = 7200 N
so the answer is "B"
A 2 kg stone is dropped from a height of 100 m. How far does it travel in the third second? take g = 9.8 m/s2
Answer:
S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m
Explanation:
CR Physical Science B (GP) 20-21 / 1 Motion
1. A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s.
What is the average speed of the object?
O 4 m/s
O
2 m/s
O 8 m/s
10 m/s
Answer:
Average speed = total distance traveled / total time traveled
V = (2 + 80) / (2 + 40) = 82 / 42 = 2 m / s
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
To solve more questions on Work, Energy and Power, visit the link below -
https://brainly.com/question/208670
#SPJ2
Which of the following statements is false?
Weight is a vector quantity
Weight is measured in newtons. N
The weight of an object is the same on the Earth and the moon
Answer:
the weight of an object is the same on earth and moon
Explanation:
bcoz weight depends on both mass and gravity
since the gravity of earth and moon is different then the weight is also different
mass doesn't change not weight
The temperature of a quantity of an ideal gas is a. one measure of its ability to transfer thermal energy to another body. b. proportional to the average molecular kinetic energy of the molecules. c. proportional to the internal energy of the gas. d. correctly described by all the statements above. e. correctly described only by the first two statements above.
Answer:
d. correctly described by all the statements above.
Explanation:
Kinetic molecular theory of gases states that gas particles exhibit a perfectly elastic collision and are constantly in motion.
According to the kinetic-molecular theory, the average kinetic energy of gas particles depends on temperature.
This ultimately implies that, the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas. Thus, an increase in the average kinetic energy of gas particles would cause an increase in the absolute temperature of an ideal gas.
Temperature can be defined as a measure of the degree of coldness or hotness of a physical object. It is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).
Generally, the temperature of a quantity of an ideal gas is;
a. a measure of the ability of an ideal gas to transfer thermal energy to another body.
b. the average kinetic energy of gas particles is directly proportional to the absolute temperature of an ideal gas
c. proportional to the internal energy of the gas.
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
A pair of butterflies reproduces and has one thousand offspring. All one thousand of the offspring have the alleles Aa. What is the most likely combination of alleles (genotype) for each parent?
Answer:
Alleles AA and aa
Explanation:
For all the offsprings of the butterflies to have the same heterozygous alleles Aa, it means the two parents have different homogyzous alleles. That is, one of the parents had the alleles AA while the other had the alleles aa. Thus, a combination of the two pairs of alleles will produce 100% Aa alleles in the offspring as seen in the image attached.
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-axis a distance d from the origin. Particle 2 carries a charge of -2q and is located on the positive x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?
Answer:
The net field will be the sum of the fields created by each charge.
where the charge Q in a position r' is given by:
E(r) = k*Q/(r - r')^2
Where k is a constant, and r is the point where we are calculating the electric field.
Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:
E₁(r) = k*3q/(r - r₁)^2
While for the other charge of -2q in the position r₂ = (d, 0, 0)
The electric field is:
E₂(r) = -k*2*q/(r - r₂)^2
Then the net field at the point r is:
E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2
E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Then if the we want to find the points r = (x, y, z) such that:
E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)
Then we must have:
0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)
Also remember that the distance between two points:
(x, y, z) and (x', y', z') is given by:
D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)
Then we can rewrite:
r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)
= √( (x + d))^2 + y^2 + z^2)
and
r - r₂ = √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)
= √( (x - d))^2 + y^2 + z^2)
Replacing that in our equation we get:
0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)
0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)
We want to find the values of x, y, z such that the above equation is true.
2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)
2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]
2*(x + d)^2 + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2
2*(x + d)^2 - 3*(x - d)^2 = 3*y^2 + 3*z^2 - 2*y^2 - 2*z^2
2*(x + d)^2 - 3*(x - d)^2 = y^2 + z^2
2*x^2 + 2*2*x*d + 2*d^2 - 3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2
-x^2 + 10*x*d - d^2 = y^2 + z^2
we can rewrite this as:
- ( x^2 - 10*x*d + d^2) = y^2 + z^2
now we can add and subtract 24*d^2 inside the parenthesis to get
- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) = y^2 + z^2
-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2
-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2
The thing inside the parenthesis is a perfect square:
-(x - 5d)^2 + 24d^2 = y^2 + z^2
we can rewrite this as:
24d^2 = y^2 + z^2 + (x - 5d)^2
This equation gives us the points (x, y, z) such that the electric field is zero.
Where we need to replace two of these values to find the other, for example, if y = z = 0
24d^2 = (x - 5d)^2
√(24d^2) = x - 5d
√24*d = x - 5d
√24*d + 5d = x
so in the point (√24*d + 5d, 0, 0) the net field is zero.
what is the main limitation of debye huckel theory
Answer:
Explanation:
For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. This is known as the Debye–Hückel limiting law.
What is the y component of a vector that is 673 m at -38o?
Answer:
D_y = 414.38m
Explanation:
D_y = D*sin(x)
D_y = 673m*sin(38°)
D_y = 414.38m
What must be the same for any two resistors that are connected in a series
Lực tương tác giữa hai điện tích điểm khi đặt trong không khí là 1,5 N. Nhúng hai điện tích đó vào môi trường điện môi có hằng số điện môi là 3 thì lực tương tác giữa chúng là bao nhiêu?
Answer:
The force is now 0.5 N.
Explanation:
Force = 1.5 N
dielectric constant , k = 3
Let the two charges are q and q' and the distance between them is r.
The electrostatic force between the two charges is given by
[tex]F \alpha \frac{ q q'}{r^2}..... (1)[/tex]
When a dielectric material is inserted between the two charges, the new force is
[tex]F' \alpha \frac{ q q'}{kr^2}..... (2)[/tex]
From (1) and (2)
F' = F/K = 1.5/3 = 0.5 N
A child is playing in a park on a rotating cylinder of radius, r , is set in rotation at an angular speed of w. The Base of the cylinder is slowly moved away, leasing the child suspended against the wall in a vertical position.
What Is the minimum coefficient of friction between the child's clothing and wall is needed to prevent it from falling .
Answer:
[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]
Explanation:
From the given information:
The force applied to the child should be at equilibrium in order to maintain him vertically hung on the wall.
Also, the frictional force acting on the child against gravitational pull is:
[tex]F_f = \mu _sN[/tex]
where,
the centripetal force [tex]F_c[/tex] acting outward on the child is equal to the normal force.
[tex]F_c= N[/tex]
SO,
[tex]F_f = \mu_s F_c[/tex]
Since the centripetal force [tex]F_c = \dfrac{mv^2}{r}[/tex]
Then:
[tex]F_f = \dfrac{ \mu_s \times mv^2}{r}[/tex]
Using Newton's law, the frictional force must be equal to the weight
[tex]F_f = W[/tex]
[tex]\dfrac{ \mu_s \times mv^2}{r} = mg[/tex]
[tex]\dfrac{ \mu_s v^2}{r} = g[/tex]
Recall that:
The angular speed [tex]\omega = \dfrac{v}{r}[/tex]
Therefore;
[tex]g = \mu_s \omega^2 r[/tex]
Making the coefficient of friction [tex]\mu_s[/tex] the subject of the formula:
[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
Two astronauts, each having a mass of 88.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.40 m/s. Treating the astronauts as particles, calculate each of the following.
a. the magnitude of the angular momentum of the system
b. the rotational energy of the system
c. What is the new angular momentum of the system?
d. What are their new speeds?
e. What is the new rotational energy of the system
Answer:
a) L = 4.75 103 kg m² / s, b) K_total = 2.57 10³ J,
c) L₀ = L_f =4.75 103 kg m² / s, d) K = 1.03 10⁴ J, K = 1.03 10⁴ J
Explanation:
a) the angular momentum is the sum of the angular momentum of each astronaut
the distance is measured from the center of the circle r = 10/2 = 5.0 m
L = 2m v r
L = 2 88.0 5.40 5.0
L = 4.75 103 kg m² / s
b) rotational kinetic energy
K = ½ I w²
As there are two astronauts, the total energy is the sum of the energy of each no.
The moment of inertia of a point mass
I = m r²
I = 88 5²
I = 2.2 10³ kg m²
the angular velocity is given by
v = w r
w = v / r
w = 5.40 / 5
w = 1.08 rad / s
the kinetic energy of the system
K_total = 2 K
K_total = 2 (½ I w²)
K_total = 2.2 10³ 1.08²
K_total = 2.57 10³ J
c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half
r = 2.5 m
I = 88 2.5²
I = 5.5 10² kg m²
for the change in angular velocity let us use the conservation of moment
L₀ = L_f
2Io wo = 2 I w
w = Io / I wo
w = 2.2 10³ / 5.5 10² 1.08
w = 4.32 rad / s
linear velocity is
v = w r
v = 4.32 2.5
K = 1.03 10⁴ J
the kinetic energy of the system is
K = 5.5 10² 4.32²
K = 1.03 10⁴ J
Diwn unscramble the word
Answer:
WIND Is what you're looking for
Explanation:
The word is WIND
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
The slope at point A of the graph given below is:
WILL MARK BRAINLIEST TO CORRECT ANSWER
RQ/PQ I think
rise/run
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắt đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây
xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

Which electromagnetic waves have the greatest frequencies?
Answer:
Gamma rays
Explanation:
Gamma rays have the highest frequency in the electro magnetic spectrum
