Answer:
500 g
Explanation:
m = Mass of water
c = Specific heat of water = [tex]1\ \text{cal/g}^{\circ}\text{C}[/tex]
[tex]\Delta T[/tex] = Temperature difference = [tex]100-20=80^{\circ}\text{C}[/tex]
Q = Heat added = 40000 cal
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow m=\dfrac{Q}{c\Delta T}\\\Rightarrow m=\dfrac{40000}{1\times 80}\\\Rightarrow m=500\ \text{g}[/tex]
The mass of water is 500 g.
Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a) what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Answer:
[tex]14400\ \text{N}[/tex], Attractive
[tex]3240\ \text{N}[/tex], Repulsive
Explanation:
[tex]q_1[/tex] = -20 μC
[tex]q_2[/tex] = 50 μC
r = Distance between the charges = 2.5 cm
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electrical force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}[/tex]
The magnitude of force each sphere will experience is [tex]14400\ \text{N}[/tex]
Since the charges have opposite charges they will attract each other.
Now the charges are brought into contact with each other so the resultant charge will be
[tex]q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}[/tex]
[tex]F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}[/tex]
The magntude of the force the spheres experience will be [tex]3240\ \text{N}[/tex]
The spheres have the same charge now so they will repel each other.
A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Acceleration due to gravity is g = 9.8 m/s2 O A. 102,282 J O B. 29,224J O C. 2982 J O D. 304.3J SUSM
Answer:
29223.6J
Explanation:
Given parameters:
Mass of Piano = 852kg
Height of lifting = 3.5m
Unknown:
Gravitational potential energy = ?
Solution:
The gravitational potential energy of a body can be expressed as the energy due to the position of a body;
G.P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the given parameters and solve;
G.P.E = 852 x 9.8 x 3.5 = 29223.6J
Use the above picture to fill in the blanks for the following statement.
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
Answer:
C + 2O ------ CO2
Explanation:
"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.
I didn't really get what you meant but this is my guess of what you meant
STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?
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Question
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of
32.7 m/s. It then flies a further distance of 44500 m, and afterwards, its velocity is 50.3 m/s. Find the
airplane's acceleration
acceleration:
.016m/s2
Calculate how much time clapses while the airplane covers those 44500 m
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49 ENG
1:45 PM
9/17/2020
Answer:
Explanation:
initial velocity u = 32.7 m /s
final velocity v = 50.3 m /s
displacement s = 44500 m
acceleration a = ?
v² = u² + 2 a s
50.3² = 32.7² + 2 x a x 44500
2530.09 = 1069.29 + 89000a
a .016 m /s²
time taken t = ?
v = u + at
50.3 = 32.7 + .016 t
t = 1100 s
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 m
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
[tex]v_x=v_o[/tex]
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal distance is calculated as a constant speed motion:
[tex]x = v_x.t[/tex]
Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:
x = 15*4 = 60
The range of the projectile is 60 m
When does a magnet induce an electric current in a wire coil?
O A. When the wire is connected to the coil
O B. When the magnet is near the coil
O C. When the magnet is moving back and forth in the coil
D. When the magnet is very strong
Answer:
B I believe
Explanation:
Which statement best describes energy and matter in a closed system? (2 po
O Energy and matter flow into and out of the system.
Energy can flow into and out of the system but matter cannot.
Energy and matter are contained within a closed system.
O There is no energy in a closed system; there is matter.
Answer:
Energy can flow into and out of the system but matter cannot.
Explanation:
In a closed system, energy can flow in and out of the system but matter cannot.
A closed system prevents double way flow of matter. A closed system conserves matter.For an isolated system, energy and matter cannot flow out of the system.
For open systems, energy and matter can flow out of the system.
Such systems are used for certain thermodynamics experiment.
Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in rad/s) of simple harmonic oscillations of this system?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Which of the following descriptions best describe a liquid
A
takes the shape and volume of its container
B
matter is made of atoms so tightly packed together that they cannot move around
C
has a definite volume, but takes the shape of its container
Answer:
C
Explanation:
1. An object in motion tends to stay in
motion because it has ___?____. (inertia or
terminal velocity)
Answer: intertia
Explanation:
A car is moving at an average speed of 20 meters per second. This is equivalent to
Answer:
44.73 MP/H or 71.98 KM/H
Explanation:
A displacement vector with a magnitude of 20. meters could have perpendicular components with magnitudes of A. 10. m and 10. m B. 12 m and 8.0 m 12 m and 16 m D. 16 m and 8.0 m
Answer:10.m and 10. M
Explanation:
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
A displacement vector with a magnitude of 20. meters can be decomposed in 2 perpendicular components.
They would form a right triangle, in which the displacement vector would be the hypotenuse (a) and the components would be the legs (b, c).
Given the magnitude of the legs, we can calculate the magnitude of the hypotenuse using the Pythagorean theorem.
[tex]c = \sqrt{a^{2} + b^{2} }[/tex]
Let's use this formula to calculate the displacement vector for each pair of legs.
A. 10. m and 10. m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(10.m)^{2} + (10.m)^{2} }= 14.1m[/tex]
B. 12 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (8.0m)^{2} }= 14.4m[/tex]
C. 12 m and 16 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(12m)^{2} + (16m)^{2} }= 20m[/tex]
D. 16 m and 8.0 m[tex]c = \sqrt{a^{2} + b^{2} } = \sqrt{(16m)^{2} + (8.0m)^{2} }= 17.9m[/tex]
A displacement vector with a magnitude of 20. m could have perpendicular components with magnitudes of C. 12 m and 16 m.
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what transition metal has 5 more protons than the halogen found in period 3?
Answer: Titanium (Ti)
Explanation:
First, each element has a unique atomic number Z, that is equal to the number of protons in the nucleus.
The halogen in period 3 is chlorine (Ch)
Chlorine's atomic number is Z = 17, this means that it has 17 protons.
Now we want to find a transition metal that has 5 more protons, then this transition metal has Z = 17 + 5 = 22
Now we can look at the periodic table and find the element with Z = 22, and if this is in the d-block, then this will be a transition metal.
The element with Z = 22 is titanium (Ti)
1. Opposite charges
O repel
attract
Answer:
attract
Explanation:
that is the answer
Answer:
Attract.
Explanation:
I took the quiz.
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:
The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?
Answer:
The value is [tex]\theta =407.3 \ radian[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]
The first time is [tex]t_1 = 1.00 \ s[/tex]
The second time [tex]t_2 = 6.10 \ s[/tex]
Generally the angular velocity is mathematically represented as
[tex]w = \int\limits {\alpha } \, dt[/tex]
=> [tex]w = \int\limits {7t + 8 } \, dt[/tex]
=> [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]
=> [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]
=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]
=> [tex]\theta =407.3 \ radian[/tex]
A 1200 N force acts on an object, resulting in an acceleration of 8.0 m/s2. What is the mass of the object?
Answer:
The answer is 150 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{1200}{8} \\ [/tex]
We have the final answer as
150 kgHope this helps you
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km
if you increase the frequency of a wave by 5x whats it’s period?
We know that Period of a wave is the inverse of its Frequency
So, Period = 1 / Frequency
From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period
Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times
Imagine you are running PE class. You run the first 1000 meters in 3 minutes and then get tired and run the last 600 meters in 5 minutes. What was your *average* speed?
Answer:
200 meters per minute
Explanation:
You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.
A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of magnitude 11.4 à 10-3 T. The magnitude of the force on 11.4 cm of this wire is:____.a) 11.4 * 10^-3 N
b) 0.130 N
c) 1.48 * 10^-2 N
d) 2.93 * 10^-3 N
Answer:
(d) 2.93 x 10⁻³ N
Explanation:
Given;
current in the wire, I = 11.4 A
angle of inclination, θ = 11.4⁰
magnetic field on the wire, B = 11. 4 x 10⁻³
length of the wire, L = 11.4 cm = 0.114 m
The magnitude of magnetic force on the wire is given by;
F = BILsinθ
F = (11.4 x 10⁻³)(11.4)(0.114)(sin 11.4°)
F = 0.00293 N
F = 2.93 x 10⁻³ N
Therefore, the correct option is "D"
What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0∘?
Answer:
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Explanation:
In order to solve this problem we will use the grating equation:
mλ = d Sin θ
where,
m = order = 3
λ = wavelength of light = ?
d = slit separation = 2 μm = 2 x 10⁻⁶ m
θ = angle = 60°
Therefore,
(3)λ = (2 x 10⁻⁶ m)Sin 60°
λ = 1.732 x 10⁻⁶ m/3
λ = 5.773 x 10⁻⁷ m = 577.3 nm
I need emergency help we only have 3 minutes left
Answer:
Option A
Explanation:
Ast the force is equal and the diayance is equal the beam is also balanced
22. What happens to the volume of a 1 kg of water when it is heated from 4oC to 6oC? A. Increases B. Decreases C. Stays the same
Answer:
a
Explanation:
..................
Starting from rest, a coin and a ring roll down a ramp without slipping. Which of the following are true:
A. The ring reaches the bottom first
B. The coin reaches the bottom first
C. The coin and the ring arrive at the same time
D. The one the reaches the bottom first is the one with the largest mass
E. The one that reaches the bottom first is the one with the largest diameter.
Answer:
The answer is D
Explanation:
The object with the heavy mass will reach first because the cause it is heaver so it will go faster
True or False. A projectile is an object that once set in motion, continues in motion by its own inertia.
The speed of a space shuttle is 10 / express this in /�
Answer:
268.22m/s
Explanation:
Given;
10mile/min to m/s
We need to convert between the two units;
1 mile = 1609.34m
60s = 1min
Now;
10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]
= 268.22m/s
A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
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Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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