Complete Question:
When a 40.0-g nugget of pure gold is heated from 49°C to 60.0°C, it absorbed 5339.01 of energy. What is
the specific heat of gold?
Answer:
Specific heat capacity, c = 12.134J/KgC
Explanation:
Given the following data;
Mass = 40g
Initial temperature, T1 = 49°C
Final temperature, T2 = 60.0°C
Quantity of heat = 5339.01J
To find the specific heat capacity of gold;
Heat capacity is given by the formula;
[tex] Q = mcdt[/tex]
Where;
Q represents the heat capacity or quantity of heat. m represents the mass of an object. c represents the specific heat capacity of water. dt represents the change in temperature.dt = T2 - T1
dt = 60 - 49
dt = 11°C
Substituting the values into the equation, we have;
[tex] 5339.01 = 40*c*11 [/tex]
[tex] 5339.01 = 440c [/tex]
[tex] c = \frac {5339.01}{440} [/tex]
Specific heat capacity, c = 12.134J/KgC
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grainsare there in the ball
Answer:
the number of grains in the ball is 274,848
Explanation:
Given that;
diameter = 0.5 mm
so radius r = 0.25 mm
first we determine the volume of the ball using the following equation;
V = 4/3×πr³
we substitute
V = 4/3×π(0.25)³
V = 0.06544 mm³
Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³
so;
Number of grains N = 0.06544 × 4,200,000
N = 274,848 grains
Therefore, the number of grains in the ball is 274,848
What forces are related to a person pushing a large crate accross the floor?
Answer:
98
Explanation:
Answer:
friction, air resistance
Explanation:
A ball rolls across a smooth surface at a steady 0.5 m/s. What is its acceleration? m/s^2?
Answer:
0m/s²
Explanation:
The acceleration of the given body whose motion has been described will be zero.
This is because the body moves with a constant velocity.
Acceleration is defined as the rate of change of velocity with time.
Since in this problem, the velocity is not changing per time, then the acceleration would be zero.
A body moving steadily implies that such body is not changing its motion by the virtue of its velocity and even direction.
What are the characteristics of a nebulae? (Select all that apply.)
end-stage of a star’s life
contain hydrogen
clouds of gas and dust
location of gas planets
needed to create a star
Answer:
B. contain hydrogen
C. clouds of gas and dust
E. needed to create a star
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
Some of the examples of stars are; Vega, Sun (closest to planet Earth), Antares, Betelgeus, Canopus, etc.
Stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He). The chronological order in which the formation of a star occur are;
1. Gravity pulls gas and dust together to form dense cores.
2. A protostar forms as mass increases.
3. Nuclear fusion begins under high pressure.
Scientists have been able to understand and discover that, gravity pulled materials (low-density cloud of interstellar gas and dust known as a nebula) together forming the planetary bodies in our solar system.
A dark nebula can be defined as an interstellar cloud that is so dense as a result of high concentration of gas and dust and as such it obscures the visible wavelengths of light from stars behind it, thus appearing completely opaque (dark patch) in front of a bright emission nebula or in regions having plenty stars.
The characteristics of a nebulae are;
I. It contain hydrogen.
II. Clouds of gas and dust
III. It is needed to create a star.
1. How many atoms are needed to form a molecule?
Answer:
Answer: 20
Explanation: cause
Explanation:
When two or more atoms link up, they create a molecule. e.g, A molecule of water is made of two atoms of hydrogen (H) and one atom of oxygen (O).
which two types of potentional energies would be types of mechancial energies
Answer:
Gravitation Potential Energy and Elastic Potential Energy
Explanation:
PE=abbreviation for potential energy
PE is a mechanical energy like KE is. There are two types of PE in high school physics (do not talk about other PE's like chemical PE).
Elastic PE is like the stored energy for example in a pole vault competition, the Elastic PE is stored in the pole before the vaulter goes over the desired height. (Energy stored in ELASTIC object ONLY) (commonly on the 11th grade regents springs are used to represent elastic PE)
Gravitational PE can be calculated through the formula PE=mgh although some people substitute the PE with a U. (ENERGY STORED IN OBJECT'S HEIGHT)
(I am doing a lab called, Density of solids and I need help with this question) (I am in HS and I picked Physics because I thought it was short for Physical Science )
4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data to relate your findings?
Answer: Location the data and observations collected in your lab guide
Explanation:
Which option is an example of a good blackbody radiator?
A. A mirror that reflects the light that strikes it
B. A classroom full of students in a school
C. A stove burner that glows red when it is turned on
D. Water that bends the light that enters it
Answer: C. A stove burner that glows red when it is turned on
Explanation:
I did the quiz
Suppose you have a string of length LaTeX: LL clamped down on both sides. Is it possible for the wavelength of a standing wave on this string to equal LaTeX: L/3L / 3
Answer:
Explanation:
For a string of length L clamped down on both ends , for fundamental mode of vibration
2 . λ / 4 = L
= λ / 2 = L
λ = 2 L
For other modes , the formula is
n . λ/2 = L where n is an integer .
λ = 2L / n
if n = 6
λ = 2L / 6
= L/3
So wavelength = L/3 is possible .
What happens to the amplitude of the resultant wave when two sound waves with equal amplitude constructively interfere?
Answer:
For two waves of equal amplitude interfering constructively, the resulting amplitude is twice as large as the amplitude of an individual wave. For 100 waves of the same amplitude interfering constructively, the resulting amplitude is 100 times larger than the amplitude of an individual wave.
At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 300 N/C. At a distance r2 from the charge, the field has a magnitude of 160 N/C. Find the ratio r2/r1.
Answer:
the ratio of r₂/r₁ is 1.37.
Explanation:
Given;
magnitude of electric field strength at a distance r₁, E₁ = 300 N/C
magnitude of electric field strength at a distance r₂, E₂ = 160 N/C
Electric field strength is given as;
[tex]E = \frac{kQ}{r^2} \\\\E_1r_1^2 = E_2r_2 ^2\\\\\frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \\\\\frac{E_1}{E_2} =( \frac{r_2}{r_1})^2\\\\\sqrt{\frac{E_1}{E_2} } = \frac{r_2}{r_1}\\\\\frac{r_2}{r_1} = \sqrt{\frac{300}{160} }\\\\\frac{r_2}{r_1} =1.37[/tex]
Therefore, the ratio of r₂/r₁ is 1.37.
Two surfaces in contact are moving slowly past each other. As the relative speed between the two surfaces in contact increases, what happens to the magnitude of their coefficient of kinetic friction
Answer:
It begins to increase
Explanation:
As the surfaces move slowly and then increase their relative speeds, the coefficient of kinetic friction begins to increase until it reaches it terminal value in which, it can no longer increase.
rank in terms of momentum
A. A 10,000 kg train car at rest
B. A 100 kg person running at 5 m/s
C. A 1200 kg car going 15 m/s
D. A 15 kg meteor going at a speed of 1000 m/s
Answer:
See the explanation below,
Explanation:
Momentum is defined as the product of mass by Velocity. In this way, we will replace the following equation in each of the cases.
[tex]P=m*v[/tex]
where:
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
A)
[tex]P = 10000*0\\P=0[/tex]
B)
[tex]P=100*5\\P=500[kg*m/s][/tex]
C)
[tex]P=1200*15\\P=18000[kg*m/s][/tex]
D)
[tex]P=15*1000\\P=15000[kg*m/s][/tex]
From higher to lower momentum
C,D,B,A
Why don't we use x-ray to communicate information?
Answer: The difficulty comes in the modulation, and in the fact that X-rays can't be transmitted via wires or fiber.
Explanation: Mark me as brainlisti
A 5 kg rock falls from an 8 m high cliff. What is the speed of the rock when it hits the ground?
Answer:
3k mph
Explanation:
don't take my answer it is wrong
a projectile is fired so that when it reaches the maximum height h it has speed vtop when the projectile is at h/2 the speed is 2vtop at what angle is the projectile fired
Answer:
Explanation:
Let the velocity of firing be u at angle θ
At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ
So , vtop = v cosθ
At height h/2
vertical component of velocity v₂
v₂² = (usinθ)² - 2 g . h/2
v₂² = u²sin²θ - gh
horizontal component of velocity at height h/2 = u cosθ
velocity at height h / 2
= √ ( u²sin²θ - gh + u² cos²θ)
Given
√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop
u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ
u²sin²θ - gh = 3 u² cos²θ
At height h , vertical component of velocity is zero
0 = u²sin²θ - 2gh
gh = u²sin²θ / 2
u²sin²θ - u²sin²θ / 2 = 3 u² cos²θ
u²sin²θ / 2 = 3 u² cos²θ
Tan²θ = 6
Tanθ = 2.45
θ = 68⁰ .
A 5.0 kg object suspended on a spring oscillates such that its position x as a function of time t is given by the equation x (t ) = A cos(ωt ), where A = 0.80 m and ω = 2.0 s-1 . What is the magnitude of the maximum net force on the object during the motion?
Answer:
The magnitude of the maximum net force on the object during the motion is 16 newtons.
Explanation:
The position of the mass, measured in meters, within the mass-spring system is:
[tex]x(t) = A\cdot \cos \left(\omega\cdot t \right)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in meters.
[tex]t[/tex] - Time, measured in seconds.
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
The acceleration function ([tex]\ddot {x} (t)[/tex]), measured in meters per square second, is obtained by deriving twice (1) in time:
[tex]\dot {x} (t) = -\omega\cdot A\cdot \sin (\omega\cdot t)[/tex]
[tex]\ddot {x} (t) = -\omega^{2}\cdot A\cdot \cos (\omega\cdot t)[/tex] (2)
And the maximum acceleration ([tex]a_{max}[/tex]), measured in meters per square second, experimented by the mass is:
[tex]a_{max} = \omega^{2}\cdot A[/tex] (3)
And the maximum net force ([tex]F[/tex]), measured in newtons, is:
[tex]F = m\cdot a_{max}[/tex] (4)
If we know that [tex]\omega = 2\,\frac{rad}{s}[/tex], [tex]A = 0.80\,m[/tex] and [tex]m = 5\,kg[/tex], then the maximum net force on the object during the motion is:
[tex]a_{max} = \left(2\,\frac{rad}{s} \right)^{2}\cdot (0.80\,m)[/tex]
[tex]a_{max} = 3.2\,\frac{m}{s^{2}}[/tex]
[tex]F = (5\,kg)\cdot \left(3.2\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 16\,N[/tex]
The magnitude of the maximum net force on the object during the motion is 16 newtons.
The magnitude of the maximum net force on the object during the motion will be 16 newtons.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
The given data in the problem is;
m is the mass of 5.0 of object kg
x (t ) = A cos(ωt ),
A is the amplitude= 0.80 m
ω is the frequency= 2.0 s-1 .
F is the magnitude of the maximum net force=?
The position of the mass is given by the equation;
[tex]x(t)= A cos( \omega t)[/tex]]
[tex]\em \dot x= - \omega A sin(\omega t) \\\\ \dot \dot x = - \omega ^2 A cos (\omega t)[/tex]
The maximum acceleration is found as;
[tex]\rm a_{max} = \omega ^2 A \\\\ \rm a_{max} = (2) ^2 \times 0.80 \\\\ \rm a_{max} =3.2 \m/sec^2[/tex]
The value of net force is found as;
[tex]\rm F= m a_{max} \\\\ \rm F= 5 \times 3.2 \\\\\ \rm F=16\ N \\[/tex]
Hence the magnitude of the maximum net force on the object during the motion will be 16 newtons.
To learn more about the force refer to the link;
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Fill in the chart with the correct values of F for each change in the system described in questions
Answer:
We know that the gravitational force F between two masses P and Q, that are separated by a distance R is:
[tex]F = G*\frac{P*Q}{R^2}[/tex]
Where G is the gravitational constant.
a) Mass P is doubled, then we have 2*P instead of P, the new force is:
[tex]F' = G*\frac{(2*P)*Q}{R^2} = 2*(G*\frac{P*Q}{R^2} ) = 2*F[/tex]
b) Now R is doubled, then instead of R, we have 2*R:
[tex]F' = G*\frac{P*Q}{(2*R)^2} = G*\frac{P*Q}{4*R^2} = G*\frac{P*Q}{R^2}*(1/4) = F/4[/tex]
c) Now we replace P by 2*P, and Q by 3*Q
[tex]F' = G*\frac{(2*P)*(3*Q)}{R^2} = 2*3*(G*\frac{P*Q}{R^2} ) = 6*F[/tex]
d) The entire mass of the system is increased by a factor of 4, then both of the individual masses are increased by a factor of 4.
Then we need to replace P by 4*P, and Q by 4*Q.
[tex]F' = G*\frac{(4*P)*(4*Q)}{R^2} = 4*4(G*\frac{P*Q}{R^2} ) = 16*F[/tex]
e) Now we replace R by R/2.
[tex]F' = G*\frac{P*Q}{(R/2)^2} = G*\frac{P*Q}{R^2/4}= 4*G\frac{P*Q}{R^2} = 4*F[/tex]
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.65 m/s. The car is a distance d away.
The bear is 32.4 m behind the tourist and running at 5.68 m/s. The tourist reaches the car safely. What is the maximum possible
value for d?
Answer:
146.27 m
Explanation:
From the question given above, the following data were obtained:
Velocity of tourist (vₜ) = 4.65 m/s
Distance travelled by tourist (dₜ) = d
Velocity of bear (v₆) = 5.68 m/s
Distance travelled by bear (d₆) = 32.4 + d
Value of d =?
Next, we shall determine the time taken for tourist and the bear to get to the car.
For the tourist:
Velocity of tourist (vₜ) = 4.65 m/s
Distance travelled by tourist (dₜ) = d
Time (tₜ) =?
vₜ = dₜ / tₜ
4.65 = d/tₜ
Cross multiply
4.65 × tₜ = d
Divide both side by 4.65
tₜ = d / 4.65
For the bear:
Velocity of bear (v₆) = 5.68 m/s
Distance travelled by bear (d₆) = 32.4 + d
Time (t₆) =?
v₆ = d₆ / t₆
5.68 = (32.4 + d) / t₆
Cross multiply
5.68 × t₆ = (32.4 + d)
Divide both side by 5.68
t₆ = (32.4 + d) / 5.68
NOTE: Both the tourist and the bear have the same time. Thus, to obtain the value of d, we shall equate both the time taken for the tourist and the bear together. This is illustrated below:
Time taken by tourist (tₜ) = time taken by the bear (t₆)
tₜ = d / 4.65
t₆ = (32.4 + d) / 5.68
tₜ = t₆
d / 4.65 = (32.4 + d) / 5.68
Cross multiply
d × 5.68 = 4.65 (32.4 + d)
5.68d = 150.66 + 4.65d
Collect like terms
5.68d – 4.65d = 150.66
1.03d = 150.66
Divide both side by 1.03
d = 150.66 / 1.03
d = 146.27 m
Thus, the maximum value of d is 146.27 m
A 110 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.220 m/s. How much work must be done on the hoop to stop it
Answer:
the work that must be done to stop the hoop is 2.662 J
Explanation:
Given;
mass of the hoop, m = 110 kg
speed of the center mass, v = 0.22 m/s
The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;
W = ΔK.E
W = ¹/₂mv²
W = ¹/₂ x 110 x 0.22²
W = 2.662 J
Therefore, the work that must be done to stop the hoop is 2.662 J
The amount of work that must be done to stop a hoop of mass 110 kg moving with s speed of 0.220 m/s is 2.662 J
Work can be defined as the ability or the capacity to perform work. Work is also the measure of energy of a body. The S.I unit of work is Joules (J)
From the question, The kinetic energy of the hoop = The energy required to stop the hoop.Formular for kinetic energy of the hoop is
EK = mv²/2.................... Equation 1
Where EK = Kinetic energy of the hoop, m = mass of the hoop, v = velocity of the hoop.
Given: m = 110kg, v = 0.220 m/s.
Substitute these values into equation 1
EK = 110(0.22)²/2
Ek = 2.662 J.
Therefore, The amount of work that must be done on the hoop to stop it is
Learn more about Work here: https://brainly.com/question/12106685
Modern psychology traces its roots back
Answer:
Psychology traces its roots back through recorded history to India, China, the Middle East, and Europe. Buddha and Confucius focused on the powers and origin of ideas. ... The ideas of Bacon and Locke contributed to the development of modern empiricism.
Explanation:
When scientists want to determine the actual age of a fossil, they use_____________
Answer:
Relative dating is used to determine a fossils approximate age by comparing it to similar rocks and fossils of known ages. Absolute dating is used to determine a precise age of a fossil by using radiometric dating to measure the decay of isotopes, either within the fossil or more often the rocks associated with it.
please genuinely help. 15 pts
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A force of 62 N acts on a 35 kg object for 10.0s. What is the objects change in momentum?
Answer:
Δp = 620 N.s
Explanation:
Newton's second law, was originally conceived as follows:[tex]F = \frac{\Delta p}{\Delta t} (1)[/tex]
In this way, a net force applied on an object, is equal to the rate of change of the momentum of the object regarding time.for m= constant, it reduces to F= m*a.
Replacing in (1) F= 62 N, and Δt = 10.0 s, we can solve for Δp (object's change in momentum) , as follows:[tex]\Delta p = F* \Delta t = 62 N* 10.0 s = 620 N*s (2)[/tex]
What needs to happen to make electrical current flow in a generator?
Two marbles, one twice as massive as the other, are dropped from the same height. When they strike the ground, how does the kinetic energy of the more massive marble compare to that of the other marble
Answer: The more massive one will have larger kinetic energy.
Explanation:
We know that when we drop an object, the acceleration of the object will be equal to the gravitational acceleration.
a(t) = -9.8m/s^2
And to get the velocity, we need to integrate over time, to get:
v(t) = (-9.8m/s^)*t + v0
Where v0 is the initial speed of the object.
You can see that the mass of the object does not affect the velocity of it.
Then when we drop two marbles of different masses from the same height, we know that the final velocity of them will be equal.
Now, we also know that the kinetic energy can be written as:
K = (m/2)*v^2
where m is the mass, and v is the velocity.
Then the kinetic energy of the marble with less mass can be written as:
k = (m/2)*v^2
And the kinetic energy for the more massive one is:
K = (M/2)*v^2
And we know that both of them have the same velocity, and M is larger than m, then we can conclude that the marble with larger mass will have larger kinetic energy.
Two 10-cm-diameter charged rings face each other, 23.0 cm apart. Both rings are charged to 40.0 nC . What is the electric field strength You may want to review (Pages 641 - 643) . For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Charge on a Van De Graaff.
The strength of the electric field is 1.10×10^4 N/C.
The data is sent to us.
r = d/2 = 0.1/2 = 0.05 m, d = 10 cm = 0.1 m, and z = 17 cm = 0.17 m
Q = 40 nC = 40×10^-9 C
A) in the middle of the two circles
Since the electric fields created by the two rings are directed in opposite directions, there is no electric strength.
B) midway down the left ring
Electric strength is equal to the sum of the electric fields created by the left and right rings.
Electric strength is equal to E leftring + E rightring, which equals 0 plus kQz/(r2 + z2).
Electric strength = k×Q×z/(r^2 + z^2)^
Electric strength = 9×10^9×40×10^-9×0.17/(0.05^2 + 0.17^2)
Electric strength= 1.10×10^4 N/C
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QUESTION 1
A box, initially at rest, has 33.3 N of force exerted on it for 2.35 s. If the box has a mass of 7.33 kg, what was its velocity at this time?
Answer:
Velocity of the box: 10.67 m/s
Explanation:
Recall that impulse is defined as the force applied times the time the force is in action. Then we can obtained the value of the impulse given to the box as:
Impulse = F * time = 33.3 N * 2.35 s = 78.225 kg m/s
then we use the momentum of the box at that precise time to find its velocity recalling that :
Momentum = mass * velocity
Then
78.225 kg m/s = 7.33 kg * v
v = 78.225/7.33 m/s = 10.67 m/s
9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en el émbolo para reducir su volumen a 35 mL, a temperatura constante? Indica la ley que aplicas.
Answer:
2.61 atm
Ley de Boyle
Explanation:
[tex]P_1[/tex] = Presión inicial = 0.96 atm
[tex]P_2[/tex] = Presión final
[tex]V_1[/tex] = Volumen inicial = 95 mL
[tex]V_2[/tex] = Volumen final = 35 mL
En este problema usaremos la ley de Boyle.
[tex]\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}[/tex]
La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.
6.A nurse pushes a woman in a wheelchair at a constant acceleration up a ramp for a period of 17.5 s. At the base of the ramp, the wheel chair is at rest, and at the top of the ramp, the
wheelchair has a velocity of 0.50 m/s. The woman and her wheelchair have a combined
mass of 65.0 kg. What is the magnitude of the net force on the woman and her wheelchair
as they acceferate up the ramp?
(A) 0.27 N
(B) 1.9 N
(C) 2.3 N
Answer:
The answer is :(B) 1.9 N
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