Answer:
Option D. 4.4 m/s²
Explanation:
The following data were obtained from the question:
Velocity (v) = 21 m/s
Radius (r) = 100 m
Centripetal acceleration (a) =.?
The centripetal acceleration of the car can be obtained as follow:
Centripetal acceleration (a) = Velocity square (v²) / radius (r)
a = v²/r
a = 21²/100
a = 441/100
a = 4.41 ≈ 4.4 m/s²
Therefore, the centripetal acceleration of the car is 4.4 m/s².
how can i solve density
Answer:
p=m÷v
Explanation:
that is the formula for it. now, using your numbers p=density,m=mass and v= volume. you just divide it.
after you measurement would be in cm/3
Answer:
You can simply find density by using formula
[tex]d = \frac{m}{v} [/tex]
where m represents mass of an object and v represents volume of an object.
hope it helps..
A hollow circular shaft made of 304 stainless steel uniformly tapers from an outer diameter of 3.0 cm to an outer diameter of 4.0 cm over a length of 2 meters. The inner diameter of 1.0 cm is constant over the length of the shaft. The shaft is subjected to an applied torque of 500 Nm. Determine the maximum shear stress in the shaft.
Answer:
maximum shear stress = τ(max) = 95.49 × 10⁶N/m²
Explanation:
given
outer diameter at one end(D₁) = 3.0cm
outer diameter at the other end(D₂) = 4.0cm
inner diameter(d₁) = 1.0cm
torque applied(T) = 500Nm
maximum shear stress will occur at lower outer diameter
the formula is τ/r = T/J
τ= T × r/J
where r is radius
T is the torque
J is the polar 2nd M of area
attached is the calculation of the question
A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 142 feet per second from an initial height of 93 feet off the ground, then the height of the projectile, h h, in feet, t t seconds after it's shot is given by the equation:
Explanation:
It is given that,
Initial velocity of the projectile, u = 142 ft/s
Initial height off the ground, [tex]h_o = 93\ feet[/tex]
We need to find the height of the projectile t seconds after its shot. It is a concept of kinematics. The equation of projectile is given by the formula as follows :
[tex]h= -16t^2+ut+h_o[/tex]
t is time in seconds
So, putting all the values we get :
[tex]h= -16t^2+128t+112[/tex]
Hence, this is the required solution.
Please help! Will give brainliest. 10 points. Show work!
Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
2x+4y
GCF for the polynomial
Answer:
[tex]\boxed{2}[/tex]
Explanation:
The GCF is the greatest common factor of the polynomial.
[tex]2x = 1, \ 2, \ x[/tex]
[tex]4y=1, \ 2, \ 4,\ y[/tex]
The greatest common factor is 2.
Given the reaction: N2(g) +2O2(g) ⇌ 2NO2(g) The forward reaction is endothermic. Determine which of the following changes would result in more reactant being produced. I. Increase NO2 II. Increase O2 III. Add a catalyst IV. Increase the temperature V. Decrease the pressure A. I and V B. II and IV C. II, III, and V D. All
Answer:
A. I and V
Explanation:
According to Le Chatelier's Principle, increasing the product side will cause the equilibrium to shift back towards the reactant side, so I is true. By the same principle, II is false.
For gases, decreasing the pressure will cause the equilibrium to shift towards the side with higher number of moles. So V is true.
The reaction is endothermic, so increasing the temperature will shift the equilibrium to the products, so IV is false. And adding a catalyst has no effect on the equilibrium, so III is false.
The changes would result in more reactants being produced will increase NO₂ and decrease the pressure. Then the correct option is A.
What is a chemical reaction?The chemical reaction is the reaction between two reactants that led to the formation of products.
The products are substances that form after the reaction. The reactants are the substances that are original materials.
Given the reaction: N₂(g) +2O₂(g) ⇌ 2NO₂(g)
The forward reaction is endothermic. Increasing the product side will cause the equilibrium to shift back towards the reactant side, so I is true.
By the same principle, II is false.
For gases, the decrease in pressure will shift the equilibrium toward the high number of moles side.
So V is true.
The reaction is endothermic, so increasing the temperature will shift the equilibrium to the products, so IV is false. Adding a catalyst has no effect on the equilibrium, so III is false.
The changes would result in more reactants being produced will increase NO₂ and decrease the pressure.
Then the correct option is A.
Learn more about chemical reactions.
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What height will the object reach? 12 points. Will give brainliest.
Answer:
12.7 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 56.7 Km/hr
Maximum height (h) =..?
First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:
Initial velocity (m/s) = 56.7 x 1000/3600
Initial velocity (m/s) = 15.75 m/s
Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:
Initial velocity (u) = 15.75 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
v = u – gt (since the ball is going against gravity)
0 = 15.75 – 9.8 × t
Rearrange
9.8 × t = 15.75
Divide both side by 9.8
t = 15.75/9.8
t = 1.61 secs.
Finally, we shall determine the maximum height as follow
h = ½gt²
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 1.61 secs.
Height (h) =..?
h = ½gt²
h = ½ × 9.8 × 1.61²
h = 4.9 x 1.61²
h = 12.7 m
Therefore, the maximum height reached by the ball is 12.7 m
The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 7 in. long and increasing at the rate of 0.3 in./s. How fast is the volume of the cube changing (in cu in/s) at that instant of time
Answer:
my big long peen for ur mom
Explanation:
In a normal use bicycle, the angular speed of the minor gear has been found to be 25 rad / s. Find the angular speed of the larger gear, knowing that the small gear has a diameter of 4 R and the large gear 10 R.
Answer:
10 rad/s
Explanation:
The gears have the same linear speed.
ω₁ r₁ = ω₂ r₂
(25 rad/s) (2R) = ω (5R)
ω = 10 rad/s
How does increasing frequency affect the crests of a wave?
They get higher.
They get closer together.
They get lower.
They get farther apart.
Answer:
they get closer together
Explanation:
A man drives his car 5 km east of his starting point then he travels 10 km south and reaches a point. From here he travels 15 km towards west and reaches another point .Finally he travels 20 km towards north. Find the net displacement of the man from where he started. plz nswer it and ell the scale or it!!!!!!!!!!!!!!!!!!!!!
Answer:
According to the question ,
Source point is A and the destination point is D
We have to find the distance of AD
From figure,
AE = AB - EB
EB = DC and ED = BC
Therefore in triangle AED
AD² = AE² + ED²
AD² = 4² + 3²
AD² = 16 + 9
AD² = 25
AD = 5
So , the man is 5km far from the starting point
Hope this helps and pls mark as brainliest :)
It is nighttime, and you dropped your goggles into a swimming pool that is 3.0m deep. If you hold a laser pointer 1.0m directly above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0m from the edge. How far are the goggles from the edge of the pool
Answer:
The distance of the goggle from the edge of the pool is 4.726 m
Explanation:
The given information are;
The depth of the swimming pool = 3.0 m deep
The height of the laser pointer above the swimming pool edge = 1.0 m
The distance from the pool edge the laser pointer enters the water = 2.0 m
The angle between the pool and the laser = ∅ = tan⁻¹(1/2) = 26.57°
Therefore, the angle of incidence to the vertical [tex]\theta_i[/tex], = θ - ∅ = 90 - 26.57° = 63.43°
By Snell's law we have;
The ratio of the sin of the angle of incidence to the sin of the angle of refraction is a a constant equal to the ratio of the refractive indices as follows;
[tex]\dfrac{sin (\theta_i)}{sin (\theta_r} =\dfrac{n_r}{n_i}[/tex]
Where:
[tex]n_r[/tex] = Refractive index of the refractive medium which is water = 1.33
[tex]n_i[/tex] = Refractive index of the incidence medium which is air = 1.00
Therefore;
[tex]\theta_r = sin^{-1} \left (\dfrac{n_i \times sin (\theta_i)}{n_r} \right) = sin^{-1} \left (\dfrac{1\times sin (63.43)}{1.33} \right) = 42.26 ^{\circ}[/tex]
We have that tan([tex]\theta_r[/tex]) = (Distance of the goggles from the point directly above the point of incidence of the beam)/(The water depth)
tan(42.26) = (The horizontal distance of the goggles from the point of incidence of the laser on the water surface)/(3.0)
∴ The horizontal distance of the goggles from the point of incidence of the laser on the water surface = 3.0 × tan(42.26) = 2.726 m
The distance of the goggle from the edge of the pool = The horizontal distance of the goggles from the point of incidence of the laser on the water surface + The distance from the edge of the water surface the laser enters the water
The distance of the goggle from the edge of the pool = 2.726 + 2 = 4.726 m
The distance of the goggle from the edge of the pool = 4.726 m.
A ball is launched horizontally at 150 m/s from a cliff. What is its horizontal velocity after 3 seconds? A. 0 m/s B. 30 m/s C. 150 m/s D. 50 m/s
Answer:
50m/s
Explanation:
Formula:
velocity= distance/ time
velocity= 150/ 3
velocity= 50 m/s
I hope this helped you :)
When a ball is launched horizontally at 150 m/s from a cliff then its horizontal velocity after 3 seconds would be 150 m/s, therefore the correct option is C.
What is Velocity?
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second. It can also be represented by the infinitesimal rate of change of displacement with respect to time. The generally considered unit for velocity is a meter per second.
The mathematical expression for velocity is given by
velocity= total displacement/Total time
Suppose an object has a total displacement of 100 meters in a total time of 25 seconds its velocity would be 4 meter/seconds
For the given problem situation when a ball is launched horizontally at 150 m/s from a cliff then its horizontal velocity after 3 seconds would be 150 m/s because the horizontal component of the velocity remains constant during a projectile motion, therefore, the correct option is C.
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Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. How many ozone molecules are present in 2.00 L of air under the stratospheric ozone conditions of 275 K temperature and 1.89 × 10−3 atm pressure?
Answer:
1.01×10^20 molecules of ozone.
Explanation:
Data obtained from the question include:
Volume (V) = 2 L
Temperature (T) = 275 K
Pressure (P) = 1.89×10¯³ atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) of ozone =.?
Using the ideal gas equation, we can obtain the number of mole of ozone as follow:
PV = nRT
1.89×10¯³ x 2 = n x 0.0821 x 275
Divide both side by 0.0821 x 275
n = (1.89×10¯³ x 2) /(0.0821 x 275)
n = 1.67×10¯⁴ mole.
Therefore the number of mole of ozone in 2 L of air is 1.67×10¯⁴ mole.
Finally, we shall determine the number of molecules present in 1.67×10¯⁴ mole of ozone.
This can be obtained as follow:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of ozone contains 6.02×10²³ molecules.
If 1 mole of ozone contains 6.02×10²³ molecules,
therefore, 1.67×10¯⁴ mole of ozone will contain = 1.67×10¯⁴ x 6.02×10²³ = 1.01×10^20 molecules.
Therefore, 1.01×10^20 molecules of ozone are present in 2 L of air.
A light bulb lights up with a potential difference of 120 V and a current of 1.5 A. What is the energy consumption of the light bulb?
Power consumption = (voltage) x (current)
Power = (120 V) x (1.5 A)
Power = 180 watts
But 1 watt = 1 Joule/second
180 watts = 180 Joule/second
Energy consumption of the light bulb =
(180) x (number of seconds it's turned on) Joules
Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s. What is the velocity of car A after the collision? Would greatly appreciate the actual answer!
Answer:
Hello There. ☆~\《--_^■^_--》\~☆ The final velocity of the car A is -1.053 m/s. For an elastic collision both the kinetic energy and the momentum of the system are conserved.
Hope It Helps!~ ♡
ItsNobody~ ☆
Given:
[tex]M_A = 281 \ kg \\\\ v_A= 2.82 \frac{m}{s}\\\\M_B = 209 \ kg \\\\ V_B = 1.72 \frac{m}{s}[/tex]
To find:
velocity=?
Solution:
Using the formula for the velocity of CAR after the collision:
[tex]\to V_A = \frac{V_A(M_A -M_B) +2 M_B V_B}{M_A+M_B}\\\\[/tex]
[tex]= \frac{2.82(281 - 209) +2 \times 209 \times 1. 72}{ 281+ 209}\\\\ = \frac{2.82(72) +718.96}{490}\\\\= \frac{203.04 +718.96}{490}\\\\= \frac{922.00}{490}\\\\= \frac{922}{490}\\\\=1.881 \ \frac{m}{s}[/tex]
Therefore, the answer is "[tex]\bold{1.881\ \frac{m}{s}}[/tex]"
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There is a Full Moon on September 14th. On which date will the New Moon occur? A.September 21st B.September 28th C.13th D.October 31st
Answer:
the answer B
Explanation:
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What far vertically will it travel before hiting the ground A. 40 m B. 30 m C. 60 m D. 50 m
Answer:
First, let's think in the vertical problem:
The acceleration will be the gravitational acceleration:
g = 9.8 m/s^2
a = -9.8 m/s^2
For the velocity, we integrate over time:
v(t) = (-9.8 m/s^2)*t + v0
Where v0 is the initial velocity, in this case v0 = 30m/s.
v(t) = (-9.8 m/s^2)*t + 30m/s
Now, for the position we integrate again over time, and get:
P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0
Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m
p(t) = (-4.9m/s^2)*t^2 + 30m/s*t
Now, the maximum vertical height is reached when:
v(t) = 0m/s = -9.8m/s^2*t + 30m/s
t = 30m/s/9.8m/s^2 = 3.06s
Now we can evaluate the vertical position in t = 3.06s
p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m
So, rounding down, the correct option is: C. 60 m
Why does fire has not any shadow?
A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s2(metre per sec.Square). Find how far the car will go before it is brought to rest?
The only thing that mkes this question inconvenient is that it uses a mixture of units ... speed in km/hr, and acceleration in m/s². You can't directly mash those together.
What's the speed when we express it in m/s ?
Speed = (90 km/hr) · (1,000 m/km) · (1 hr/3,600 sec)
Speed = (90 · 1,000 · 1 / 3,600) · (km-m-hr / hr-km-sec)
Speed = 25 m/s
OK great !
-- The car is traveling at 25 m/s when the brakes are applied.
-- The brakes slow it down by 0.5 m/s every second.
-- So it takes (25/0.5) = 50 seconds to stop the car.
-- During that time, the car's average speed is (1/2)·(25m/s + 0) = 12.5 m/s .
-- Moving at an average speed of 12.5 m/s for 50 sec, the car travels
(12.5 m/s) · (50 s) = 625 meters
an attempt to estimate the height of a tree the Shadow of an upright metre rule was found to be 25 cm and the length of the Shadow of the tree was 7 m what is the height of the tree
Answer:
The actual height of the tree is 28 m
Explanation:
The given information are;
The length of the shadow of an upright meter rule = 25 cm
The actual height of the meter rule = 100 cm
The length of the shadow of the tree = 7 m
The actual height of the tree = h
We have
[tex]\dfrac{The \ length \ of \ the \ shadow \ of \ an \ upright \ metre \ rule}{The \ actual \ height \ of \ the \ metre \ rule} = \dfrac{The \ length \ of \ the \ shadow \ of \ the \ tree}{The \ actual \ height \ of \ the \ tree}[/tex]Which gives;
[tex]\dfrac{25 \ cm}{100 \ cm} = \dfrac{7 \ m}{The \ actual \ height \ of \ the \ tree}[/tex]
Therefore;
[tex]The \ actual \ height \ of \ the \ tree = 7 \ m \times \dfrac{100 \ cm}{25\ cm} = 7 \ m \times 4 = 28 \ m[/tex]
That is the actual height of the tree = 28 m.
Consider that each tick mark represents 1 mile on the vector grid, use the Gizmo to simulate the following scenario: Imagine that a hiker walked 6 mi SE and then 4 mi N. Find the sum of the vectors. Which of the following best describes the general direction that the blue line points?
a) south
b) west
c) east
d) cannot be determined from this information
Answer:
The best description of the general direction that the blue line points is East direction
Answer:
c. east
Explanation:
Which of the following are double-displacement reactions? multiple answers
A. 2H2(g) + O2(g) -> 2H2O(g)
B. Zn(s) + 2HCI(aq) -> H2(g) + ZnCI2(s)
C. Ba(OH)2(aq) + 2HCIO4(aq) -> Ba(CIO4)2(aq) + 2H2O(I)
D. HNO3 (aq) + KOH (aq) -> H2O(I) + KNO3 (aq)
Answer:
C and D
Explanation:
A: is a simple composition
B: is a single replacement
C: C is a double displacement
D: is a double replacement
Answer:
The answers are C abd D.
Explanation:
Here,
A no. Is a combination or analysis chemical reaction.
B no.is a single displacement reaction as cl2 goes to zn.
C no. Is double displacement reaction.
D no. Is a double displacement reaction.
Hope it helps.....
2.0 MW is to arrive at a large shopping mall over two 0.100Ω lines. Estimate how much power is saved if the voltage is stepped up from 120 V to 1200 V and then down again, rather than simply transmitting at 120 V. Assume the transformers are each 99% efficient.
Answer:
55 MW.
Explanation:
So, we are given the following data or parameters Below;
=> "2.0 MW is to arrive at a large shopping mall over two 0.100Ω line.
=> " the voltage is stepped up from 120 V to 1200 V and then down again, rather than simply transmitting at 120 V."
=> "Assume the transformers are each 99% efficient."
STEP ONE: determine the current in the transmission line.
Output Current, i = Power/ voltage.
Output Current,i = (2.0 MW × 10^6 W/ 1 MW)/120.
Output Current,i = 1.66 × 10^4 A.
Therefore, current in the transmission line = output voltage × output current/99% × line voltage.
= 120 × 1.66 × 10^4/ 99% × 1200 = 1.68 × 10^3 A.
STEP TWO: determine the power loss in the two lines.
Power loss = i^2 × Resistance
Power loss = (1.66 × 10^4 )^2 × 0.1 × 2 = 5.5 × 10^7 watt.
STEP THREE: determine the power generated.
Power generated = 2 × 10^6 +5.5 × 10^7.
Power generated = 5.57 × 10^7 watt.
STEP FOUR: determine the step down transformer power.
= 2 × 10^6/99% = 2.02 × 10^6.
Thus, 2.02 × 10^6 + 5.57 × 10^5 = 2.58 × 10^6 watt.
STEP FIVE: Determine the total power and the saved power.
Total power = 2.58 × 10^6/ 99%= 2.6 × 10^6.
Saved power = 5.7 × 10^7 - 2.6 × 10^6 .
Conversion to MW gives saved power = 55 MW.
=
A board used to pry a boulder loose and roll it over is an example of which of these simple machines?
A.
inclined plane
B.
wheel and axle
C.
lever
D.
pulley
Answer:
C. lever
Explanation:
Any sort of prying tool is an example of a lever.
Answer:
A lever
Explanation:
Observing neutrinos from the Sun is an important way to check the fusion rate, but it can be very tough to build a machine that can detect these particles. What trouble did the most recent experiment, Borexino, have to overcome
Answer:
The trouble that the most recent experiment, Borexino, have to overcome was that
neutrinos hardly interact with matter and so radioactive decay of ant material inside the detector could look exactly like a neutrino interaction too
Which is one piece of evidence of seafloor spreading?
a. fossil material
b. continent material
c. drilled core samples
d. ocean water samples
Answer:
A. Continent material
Explanation:
Eruptions of molten material, which results in continent material is a piece of evidence of seafloor spreading.
Continent material is one piece of evidence of seafloor spreading. Option b is correct.
What is seafloor spreading?Seafloor spread is a phenomenon that happens along mid-ocean ridges, when new oceanic crust is generated by volcanic activity and then progressively travels away from the ridge.
The geological activity of seafloor spreading occurs when tectonic plates split apart.
Continent material is one piece of evidence of seafloor spreading.
Hence, option b is correct.
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Supposing we launched a very fast dart from the Space Shuttle, pointed in some direction away from any planet, so that it could travel beyond the solar system. What would it be most likely to hit first after traveling outward for a while
If we launched a very fast dart from the Space Shuttle in a direction away from any planet, Assuming the dart is traveling at an extremely high speed (close to the speed of light), it might eventually encounter a cloud of interstellar gas or dust.
What does the space contain?Space is mostly empty, consisting of a vacuum with very low density. However, space also contains various astronomical objects, such as stars, planets, moons, asteroids, comets, and other celestial bodies.
There is also interstellar gas and dust scattered throughout space, which can sometimes form clouds that give birth to new stars. Additionally, space is permeated by electromagnetic radiation, including cosmic rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves.
Space also contains various forms of energy, including dark energy, which is believed to be responsible for the accelerating expansion of the universe, and dark matter, which is thought to make up the majority of the matter in the universe but cannot be directly observed. Overall, space is a vast and complex environment that continues to fascinate astronomers and scientists.
Here in this question,
Therefore, if launched a very fast dart from the Space Shuttle, is pointed in some direction away from any planet, it could travel beyond the solar system. it would be most likely to hit a cloud of interstellar gas or dust. first after traveling outward for a while.
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This is a question on my physics test :)
Can any1 help me? if you can, please give an explanation
Answer:
119.6 J/Kg°C
Explanation:
Data obtained from the question include:
Mass of substance (ms) = 170 g
Initial temperature of substance (Ts) = 120 °C
Volume of water = 200 mL
Initial temperature of water (Ts) = 10 °C
Temperature of the mixture (T2) = 12.6 °C
Density of water = 1 g/mL
Specific heat capacity of water (Cw) = 4200J/Kg°C
Specific heat capacity of substance (Cs) =..?
Next, we shall determine the mass of water. This can be obtained as follow:
Volume of water = 200 mL
Density of water = 1 g/mL
Mass of water =..?
Density = mass /volume
1 = mass /200
Cross multiply
Mass of water = 1 x 200
Mass of water = 200 g
Convert 200 g of water to Kg
Mass of water = 200/1000 0.2 Kg
Mass of water = 0.2 Kg
Now, we obtained the specific heat capacity of the substance using the following formula:
MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0
Mass of water = 0.2 Kg
Initial temperature of water (Ts) = 10 °C
Specific heat capacity of water (Cw) = 4200J/Kg°C
Temperature of the mixture (T2) = 12.6 °C
Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg
Initial temperature of substance (Ts) = 120 °C
Specific heat capacity of substance (Cs) =..?
MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0
0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0
840(2.6) + 0.17Cs(– 107.4) = 0
2184 – 18.258Cs = 0
Rearrange
2184 = 18.258Cs
Divide both side by the coefficient of Cs i.e 18258
Cs = 2184/18.258
Cs = 119.6 J/Kg°C
Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C
A scientist wants to measure the relationship between humidity in the
summer and the number of home runs in baseball. Which of the following is a
testable question about this topic?
A. Are home runs more exciting during the summer?
O B. Should fans like the players who hit the most home runs?
O C. Are home runs more fun to watch than normal hits?
D. Are there more home runs during the more humid months of the
summer?
Answer:
D
Explanation:
The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the summer?'
Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.
Only option D raises a valid question that is relevant to the aim of the investigation.