Answer:
The equilibrium temperature is 45°C
Explanation:
Given;
mass of block of ice, m₁ = 10 kg
mass of water, m₂ = 10 kg
initial temperature of the ice, t₁ = 0.0°C
initial temperature of water, t₂ = 90.0°C
let the equilibrium temperature = T
Apply the principles of mixture;
Heat lost by the water = heat gained by the ice
m₂c(90 - T) = m₁c(T - 0)
m₂(90 - T) = m₁(T - 0)
10(90 - T) = 10(T)
90 - T = T
90 = 2T
T = 90 / 2
T = 45°C
Therefore, the equilibrium temperature is 45°C
A car transports its passengers between 3 buildings. It moves from the first building to the second building, 4.76km away, in a direction 37° north of east. It then moves from second building to the third building in a direction 69° west of north. Finally, it returns to the first building, sailing in a direction 28° east of south. Calculate the distance between (a) the second and third buildings and (b) the first and third buildings.
Answer:
1) a = 6.14 km
2) b = 4.69 km
Explanation:
Let the first building be A, second building be B and third building be C.
Now, bearing of A = 4.76 km in a direction 37° north of east
Bearing of B = 69° west of north
Bearing of C = 28° east of south
Thus if this 3 points form a triangle, we will have the following angles;
Angle at point A = 28 + (90 - 37) = 81°
Angle at point B = 28 + (90 - 69) = 49°
Angle at point C = 180 - (81 + 49) = 50°
Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;
Using sine rule, we can find "a".
a/sin 81 = 4.76/sin 50
a = 6.14 km
B) distance between first and third building is AB in the triangle depicted by "b".
Similar to the first problem, we will use sine rule again.
b/sin49 = 4.76/sin 50
b = 4.69 km
Which statement describes oneScientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.
A purple ball in the center surrounded by overlapping concentric black ovals, each with a small green ball on each of the 6 ovals.
What experimental evidence led to the development of this atomic model from the one before it? feature of Rutherford’s model of the atom?
Answer:
el continuo movimiento
Explanation:
If your mass is 72 kg, your textbook's mass is 3.7 kg, and you and your
textbook are separated by a distance of 0.33 m, what is the gravitational
force between you and your textbook? Newton's law of gravitation is
Gm;m2 The gravitational constant G is 6.67 x 10-11 N·m2/kg2.
F gravity
O A. 4.94 x 10-7N
B. 2.45 x 103 N
O C. 5.38 x 10-8N
O D. 1.63 x 10-7N
Answer:
The answer is 1.63 x 10 -7 N
Explanation:
An appropriate solution is 1.63 x 10 -7 N
Gravitational pressure method: F=G{frac{m_1m_2}{r^2}}
Newton's law of gravitation is Gm;m2
The gravitational regular G is 6.67 x 10-11 N·m2/kg2.
What is gravitational pressure?
Definitions of gravitational pressure. The force of attraction among all masses inside the universe; especially the appeal of the earth's mass for bodies close to its floor.
What law is Newton's regulation of gravitation?Newton's regulation of typical gravitation states that two bodies in an area pull on every other with a force proportional to their hundreds and the space among them. For large objects orbiting each other—the moon and Earth, for instance—means they without a doubt exert important force on each other.
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ANSWET ASAP
Which term is the rate at which work is done?
energy
power
joules
force
Answer:
power
Explanation:
Answer:
power
Explanation:
A train travels 6.1m/s (E)The train’s destination is 134 km(E), how long until it arrives?
Answer:
The time taken is [tex]t = 6.102 \ hours [/tex]
Explanation:
From the question we are told that
The velocity of the train is [tex]v_t = 6.1 \ m/s[/tex]
The distance covered by the train is [tex]D = 134 \ km = 134000 \ m[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{ 134000}{ 6.1 }[/tex]
=> [tex]t = \frac{ 134000}{ 6.1 }[/tex]
=> [tex]t = 21967.21 \ sec [/tex]
Converting to hours
[tex]t = 21967.21 = \frac{21967.21 }{3600} = 6.102 \ hours [/tex]
A bird has a mass of 0.8 kg. Calculate the weight of the bird.
Answer:
the mass and wieght means same
You want to know how tall a cliff is, so you drop a rock off the edge. You estimate that
the rock has a mass of about 15 kg. You hear a splash 1.00 seconds later. How tall is the
cliff?
Analysing the Question:
Neglecting air resistance
We are given:
mass (m) = 15 kg
time taken by the rock to reach the water (t) = 1 second
height of the cliff (h) = h m
initial velocity of the rock (u) = 0 m/s [ the rock was 'dropped' and not 'thrown' downwards]
acceleration of the rock due to gravity (a) = 9.8 m/s²
Finding the height to the cliff:
from the seconds equation of motion
s = ut + 1/2*at²
replacing the variables
h = (0)(1) + 1/2 * (9.8) *(1)²
h = 4.9 m
Therefore, the cliff is 4.9 m high
A skater glides off a frozen pond onto a patch of ground at a speed of 1.8 m/s. Here she is slowed at a constant rate of 3.00 m/s. How fast is the skater moving when she has slid 0.37 m across the ground?
Answer:
1.01 m/s
Explanation:
Given
initial speed, u = 1.8m/s
acceleration, a = -3.00m/s² (it's negative because the skater slowed down)
distance, s = 0.37m
Required
Determine the final velocity (v)
This will be solved using the following equation of motion
[tex]v\² = u\² + 2as[/tex]
Substitute values for u, a and s
[tex]v\² = 1.8\² + 2 * -3 * 0.37[/tex]
[tex]v\² = 3.24 - 2.22[/tex]
[tex]v\² = 1.02[/tex]
Take square roots
[tex]v = \sqrt{1.02[/tex]
[tex]v = 1.00995049384[/tex]
[tex]v = 1.01 m/s[/tex] (Approximated)
The final speed of the skater when she has slid 0.37 m is 1.0 m/s.
The given parameters:
Speed of the skater, v = 1.8 m/sacceleration of the skater = 3.0 m/s²Displacement of the skater = -0.37 mThe final speed of the skater when she has slid 0.37 m is calculated as follows;
[tex]v_f^2 = v_i^2 + 2ad\\\\v_f ^2 = 1.8^2 + 2(3)(-0.37)\\\\v_f^2 = 1.02\\\\v_f = \sqrt{1.02} \\\\v_f = 1.0 \ m/s[/tex]
Thus, the final speed of the skater when she has slid 0.37 m is 1.0 m/s.
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I don’t know what this is
Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to make the comparison?
Answer:
u need to make sure that comparison is = to shapes and then find the shapes sizes and add them
Which of the following best describes the accuracy of a measurement?
A.
the precision of a measuring tool
B.
the amount of uncertainty involved in a measurement
C.
the degree to which the measurement approaches the quantity's true value
D.
the degree to which repeated measurements agree
Answer:
l think the answer is C. part
The degree to which the measurement approaches the quantity's true value
is what best describes the accuracy of a measurement.
When a quantity is measured, the value is usually unknown by the individual
taking the measurement . This should however be done with a very good
tool/equipment to eliminate inaccuracy.
The accuracy of a quantity is usually measured by how close the difference
is between the measurement taken and the actual value of the quantity.
The different in value gotten should always be very small.
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A box is sliding along a frictionless surface and gets to a ramp. Disregarding friction, how fast should the box be going on the ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (Use g = 9.8 m/s2.)
Answer:
7.0 m/s
Explanation: I just did it
A student wants to determine the density of titanium. The student measures the mass of a solid chunk of titanium, but when he drops it into the water to determine the volume, some water splashes out of the cylinder. Will this cause the calculated density to be higher or lower than it should be? Explain!
Answer:
Lower than it should be.
Explanation:
Hello.
In this case, since the density is computed by:
[tex]\rho =\frac{m}{V}[/tex]
And we obtain the volume of the solid by substracting the mass of the water and the solid minus the mass of water:
[tex]V_{solid}=V_{solid\ with \ water}-V_{water}[/tex]
If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.
Regards.
A driver translation a speed of 115km/hr received a text message on his mobile phone how far is he ,in kilometers,20s later from when he received the text message?
Answer:
S=d/t.
distance is= speed multiplied by the time
D= 115 multiplied by 20
the answer is 2300km
Driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.
To determine the distance the driver travels 20 seconds after receiving the text message, we need to calculate the distance covered during that time interval.
First, we convert the driver's speed from km/h to m/s for consistent units. The conversion factor is 1 km/h = 0.27778 m/s.
Driver's speed = 115 km/h × 0.27778 m/s = 31.944 m/s
Next, we use the formula for distance traveled:
Distance = Speed × Time
Distance = 31.944 m/s × 20 s = 638.88 meters
Therefore, the driver will be 638.88 meters away from the point where he received the text message after 20 seconds.
To convert this distance to kilometers, we divide by 1000:
Distance = 638.88 m ÷ 1000 = 0.63888 km
So, the driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.
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What is the best description of the relationship between emerging scientific ideas and open-mindedness?
HURRY !
Answer: C ON EDGE 2021
Explanation:
Different satellites orbit the earth with a vast range of altitudes, from just a couple hundred km, all the way to tens of thousands of km above the surface. The international space station (ISS) is in a low earth orbit, just 400km above the surface (you can see it with the naked eye at sunset and sunrise as a bright, moving dot). At this altitude, the acceleration due to gravity has a value of 8.69.
A. What is the speed of the ISS?
B. What is the orbital period (T) of the ISS in minutes?
Answer:
Find the answer in the explanation.
Explanation:
Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.
A.) Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where orbital speed = sqrt ( g × r )
Orbital speed = sqrt ( 6800 × 8.69 )
Orbital speed = 243.09 km/h
B.) Orbital period acceleration to Kepler third law of planetary motion state that:
The square of the period is directly proportional to the cube of the radius.
T^2 = (4π^2 /GM) r^3
T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3
T^2 = (4π^2/ 4.002^36) × 6800^3
T^2 = 9.865×10^-36 × 6800^3
T^2 = 3.10 × 10^-15
T = 5.57 × 10^-8 hours.
Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where:
Orbital speed = sqrt ( g × r )Orbital speed = sqrt ( 6800 × 8.69 )Orbital speed = 243.09 km/hThe speed of the ISS is 243.09Km/h.
Part B:
The orbital period (T) of the ISS in minutes is :
T^2 = (4π^2 /GM) r^3T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3T^2 = (4π^2/ 4.002^36) × 6800^3T^2 = 9.865×10^-36 × 6800^3T^2 = 3.10 × 10^-15T = 5.57 × 10^-8 hours.Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
The orbital period (T) of the ISS in minutes is 3.4 × 10^-6 minutes.
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An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?
Answer:
[tex]a=2.8\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity of an object, u = 22 m/s
Final velocity of an object, v = 36 m/s
Time, t = 5 s
It can be assumed to find the average acceleration of the object instead of average velocity.
The change in velocity per unit time is equal to average acceleration of an object. It can be given by :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2[/tex]
So, the acceleration of the object is [tex]2.8\ m/s^2[/tex].
What is Newton's First Law of Motion? (2 points)
A For every action there is an equal and opposite reaction.
B An object in motion or at rest will stay that way unless acted on by an unbalanced force.
C The smaller the mass of an object, the greater the acceleration of that object when a force is applied.
D Every force is said to be unbalanced when the force acting against it is less.
Answer:
B
Explanation:
B is Newton's first law of motion, I got this in class and that was the right answer.
Good Luck!
The number of waves in a given time period is called the ________________ .
Answer:
frequency.
......... i need more characters so
Can someone please help, ty!!
(Will mark brainliest)
Answer:
graph
noun
a diagram representing a system of connections or interrelations among two or more things by a number of distinctive dots, lines, bars, etc.
Mathematics.
a series of points, discrete or continuous, as in forming a curve or surface, each of which represents a value of a given function.
Also called linear graph . a network of lines connecting points.
a written symbol for an idea, a sound, or a linguistic expression.
verb (used with object)
Mathematics. to draw (a curve) as representing a given function.
to represent by means of a graph.
Explanation:
that is what i found
You made a peanut butter and jelly sandwich for lunch and ate it. What did your actions cause?
A. Caused a physical change when digesting the food
B. Caused a physical change when chewing the sandwich.
C. Caused a chemical change to the peanut butter and jelly.
O D. Performed a chemical reaction by mixing the peanut butter and jelly together.
Answer:
C
Explanation:
Chemical energy is in the food you eat and It fuels your body
Answer: B
Explanation:
When you chew it, you're changing the physical state it's in.
How much kinetic energy does a 1 kg ball
traveling at 5 m/s have?
Answer:
Okay, first off, the formula for Kinetic Energy is:
KE = 1/2(m)(v)^2
m = mass
v = velcoity (m/s)
Using this formula, we can then calculate the kinetic energy in each scenario:
1) KE = 1/2(100)(5)^2 = 1,250 J
2) KE = 1/2(1000)(5)^2 = 12,500 J
3) KE = 1/2(10)(5)^2 = 125 J
4) KE = 1/2(100)(5)^2 = 1,250 J
A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 V.
Required:
a. Which point, a or b, is at the higher potential?
b. Calculate the value of £.
c. A negative point charge q = —0.200 μC is moved from 5 to a. Calculate the work done on the point charge by the electric field.
Answer:
a)
b is at higher potential
b)
E = 800 V/m
c)
W = -48.0 μ J
Explanation:
Given:
electric field has magnitude E and is directed in the negative x direction
potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) = 240 V
a)
Direction of the electric field is in the negative x direction so it is away from the positive charge and the more distance away from direction, the lesser the potential gets. So b is at the higher potential.
b)
When the magnitude of electric field is constant then the potential difference between two points in the field is given by:
Vab = Ed
where
Vab is the potential difference between two points a and b in the field
E is electric field magnitude
d is the distance between two points
Compute distance between two points a and b
d = 0.90 - 0.60
d = 0.30
d = 0.3 m
Since potential difference between two points in the field is given which is 240 V
Compute E:
The above formula becomes:
Vab = Ed
E = Vab/d
= 240/0.3
E = 800 V/m
c)
The negative charge moved from the higher potential to low so work done on the point charge by electric field is negative.
W = Fd
Electric field = E = F/q
where
E is electric field magnitude
F is electric force on q
q is point charge magnitude
The charge q is either positive or negative and when charge is positive the directions of E and F are same otherwise opposite in case of negative charge.
So to calculate the work done on the point charge by the electric field:
E = F/q
F = Eq
so
W = Fd = Eqd
Now putting the values:
E = 800 V/m
q = - 0.200 μC
d = 0.3 m
W = Eqd
W = 800 * - 0.200 * 0.3
W = 800 * - 0.2 * 10⁻⁶ * 0.3
W = - 48 * 10⁻⁶
W = - 4.8 * 10⁻⁵ J
W = -48.0 μJ
a. The point which is at a higher potential is point a.
b. The magnitude (E) of the uniform electric field is equal to 800 V/m.
c. The work done on the point charge by the electric field is equal to -48 microjoules.
Given the following data:
Point a (at x) = 0.60 meterPoint b (at x) = 0.90 meterPotential difference = 240 Volts.Point charge = -0.200 uC = [tex]0.200 \times 10^{-6}\;C[/tex]a. Since the uniform electric field with a magnitude (E) is directed in the negative x direction, the point at a higher potential is point a because the electric field is not close to the positive charge and it is further away from point b by 0.90 meter. Also, the potential difference of an electric field is inversely proportional to the square of the distance from a point charge.
b. To calculate the magnitude (E) of the uniform electric field:
At a constant potential difference, the magnitude (E) of this electric field is given by the formula:
[tex]V_{ab} = Ed[/tex]
Where:
V_{ab} is the potential difference between point a and b.E is the magnitude the electric field.d is the distance between point a and b.First of all, we would determine the distance (d).
[tex]d = d_b - d_a\\\\d = 0.90 -0.60[/tex]
Distance, d = 0.30 meter.
Substituting the given parameters into the formula, we have;
[tex]240 = 0.3E\\\\E=\frac{240}{0.3}[/tex]
E = 800 V/m
c. To calculate the work done on the point charge by the electric field:
Mathematically, the work done on a point charge in an electric field is given by the formula:
[tex]W = Fd = Eqd\\\\W = 800 \times (-0.200 \times 10^{-6}) \times 0.3\\\\W = -240 \times 0.200 \times 10^{-6}\\\\W = -48\times 10^{-6}\;Joules[/tex]
Note: 1 microjoules = [tex]1\times 10^{-6}[/tex] Joules
Work done = -48 microjoules.
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What best describes the dropping height of a ball that bounced back up to a height of 45 centimeters?
Less than 45 centimeters, as the ball transforms some of its thermal energy into potential energy
Greater than 45 centimeters, as the ball transforms some of its thermal energy into potential energy
Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy
Greater than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy
Answer:C:Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy
Less than 45 centimeters, as the ball transforms some of its potential energy into thermal energy and sound energy.
Although the initial energy (potential energy is preserved), the energy of deformation as the ball strikes a surface creates energy dissipation in the form of frictional heat and audible sound energy.
Every time the ball bounces, its height will be less than its previous height.
Explanation:
Answer:
C is incorrect I got it wrong on my test don't go with C
Explanation:
A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?
A. 55.7 newtons
B. 5,371.5 newtons
C. 273.5 newtons
D. 547 newtons
Answer:
a
Explanation:
its correct
Which of the following statements must be true?
a. The velocity and acceleration are in the direction of the resultant force.
b. The velocity could be in any direction, but the acceleration is in the direction of the resultant force.
c. The velocity and acceleration could be in any direction.
d. None of these choices.
Answer:
your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force
Please Help!!
A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?
Answer:
h = 219.96 m
Explanation:
Speed of the stone with which it was thrown horizontally, v = 10 m/s
We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.
It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0
Using second equation of motion,
[tex]d=u't+\dfrac{1}{2}at^2[/tex]
Put u'=0 and a=g
So,
[tex]d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m[/tex]
So, the height of the cliff is 219.96 m.
One strategy in a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second one at low angle times to arrive before or at the same time as the first one . Assume both snowballs are throw with a speed of 33.2 m/s. The first o e is thrown at an angle of 57 degrees with respect to the horizontal. At what angle should the second snowball be throw to arrive at the same point as the first? Answer in units of degrees
Answer:
33°
Explanation:
We are given;
Speed at which both snowballs are thrown; v = 33.2 m/s
Angle at which snowballs are thrown with respect to the horizontal; θ = 57°
Now,we want to find out the angle at which the second snowball should be thrown in order to arrive at the same point as the first.
To calculate this angle, we will use complementary angle concept.
Now, because the target is in the same place, there will be two launch angles that will make the snow ball to be placed on the target.
The is calculated from;θ1 = 45° - (57° - 45°) = 33°
A thick conducting spherical shell, with inner radius R1 and outer radius R2, is in electrostatic equilibrium. The area inside and outside the spherical shell is vacuum. The net charge on the spherical shell is Q. Note: for parts (a)-(c), describe the direction of the electric field assuming that Q is positive.
(a) Find the electric field outside the shell, at a distance d > R2 from the center of the shell.
(b) Find the electric field in the shell material, at a distance d from the center of the shell where R1 < d < R2.
(c) Find the electric field at a distance d < R1 from the center of the shell, i.e. in the empty space inside the shell.
(d) How much charge is on the inner surface of the conducting shell
Answer:
a) E = k Q / r² , k = 1 / 4πε₀, b) E=0, C) E=0, d) Q=0
Explanation:
To find the electric field in this exercise let's use Gauss's law
Ф = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
Let's use a sphere as a Gaussian surface, in this case the field lines and the radii of the spheres are parallel so the scalar product is reduced to the algebraic product, let's carry out the integral
E A = q_{int} /ε₀
The area of a sphere is
A = 4π r²
we substitute
E (4π r²) =q_{int} /ε₀
a) The field is requested outside the spherical shell
d> R₂
in this case the charge inside the Gaussian surface is
q_{int} = Q
E = Q / (4π r²) ε₀
E = k Q / r²
k = 1 / 4πε₀
b) In the second case, the field inside the spherical casing is requested.
As the surface is metallic, the charge is located on the surface of it, the electrons repel each other. So the charge inside is zero
q_{int} = 0
E = 0
c) The field in the interior region of the shell, all the charge is external, therefore the internal charge is zero
q_{int} = 0
E = 0
d) as the shell is conductive, the electrons can move freely, which is why it moves as far away as possible between them, this means that everything is located as close as possible to the outer surface of the conductor. Consequently, THERE ARE NO ELECTRONS ON THE INTERNAL SURFACE
Q_internal =0
A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil
Answer:
Δϕ = -2.89 x 10^-4 Wb
Explanation:
Given that A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil
Let ΔB = Change in magnetic field
A = cross sectional area = 8.5 cm^2
t = time = 12ms
Flux change = Δϕ = (ΔB*A) / t
Δϕ = [(0.02 - 0.06)x(8.5 x 10^-2)^2 ] / 0.012
Δϕ = (-0.04 x 7.225^-3) / 0.012
Δϕ = -2.89 x 10^-4 Wb
Therefore, the magnitude of the total change in the external magnetic flux enclosed by the coil is -2.89 x 10^-4 Wb