Answer:
Part a)
[tex]F_v = 4.28 N[/tex]
Part B)
[tex]L = 1.02 m[/tex]
Part C)
[tex]v = 1.25 m/s[/tex]
Explanation:
Part A)
As we know that ball is hanging from the top and its angle with the vertical is 20 degree
so we will have
[tex]Tcos\theta = mg[/tex]
[tex]T sin\theta = F_v[/tex]
[tex]\frac{F_v}{mg} = tan\theta[/tex]
[tex]F_v = mg tan\theta[/tex]
[tex]F_v = 1.2\times 9.81 (tan20)[/tex]
[tex]F_v = 4.28 N[/tex]
Part B)
Here we can use energy theorem to find the distance that it will move
[tex]-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2[/tex]
[tex](-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2[/tex]
[tex](-3.5 + 2.54)L = - 0.98[/tex]
[tex]L = 1.02 m[/tex]
Part C)
At terminal speed condition we know that
[tex]F_v = mg[/tex]
[tex]bv^2 = mg[/tex]
[tex]2.5 v^2 = 3.9[/tex]
[tex]v = 1.25 m/s[/tex]
A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her moment of inertia decreases by 1/2. Her angular speed increases by a factor of
Answer:
the angular velocity increase by a factor of 2
Explanation:
using the law of the conservation of the angular momentum
[tex]L_i = L_f[/tex]
where [tex]L_i[/tex] is the inicial angular momentum and [tex]L_f[/tex] is the final angular momentum.
also, the angular momentum can be calculated by:
L = IW
where I is the inertia momentum and the W is the angular velocity.
so:
[tex]I_i W_i = I_fW_f[/tex]
we know that [tex]I_f = \frac{1}{2}I_i[/tex] then,
[tex]I_iW_i = \frac{1}{2}I_iW_f[/tex]
solving for [tex]W_f[/tex]:
[tex]W_f = 2W_i[/tex]
When a figure skater pulls her arms in while spinning, her moment of inertia decreases and her angular speed increases. In the provided case, with the moment of inertia decreasing by half, the angular speed will double.
Explanation:When a figure skater is spinning with her arms outstretched, and she pulls her arms in close to her body, her moment of inertia decreases. According to the law of conservation of angular momentum, if the moment of inertia of a spinning object decreases, its angular speed must increase to keep the angular momentum constant. In this case, since the skater's moment of inertia decreases by half, her angular speed will correspondingly double, or increase by a factor of 2.
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You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and the collision takes about 20.0 ms (2.00 x 10^-2 s). a) What is the change in momentum for the blob of clay?
Answer:0.25 kg-m/s
Explanation:
Given
mass of blob [tex]m=50 gm [/tex]
initial velocity [tex]u=-5 m/s\ \hat{i}[/tex]
time of collision [tex]t=20 ms[/tex]
we know Impulse is equal to change in momentum
initial momentum [tex]P_i=mu[/tex]
[tex]P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s[/tex]
Final momentum [tex]P_f=50\times 10^{-3}v[/tex]
[tex]P_f=0[/tex] as final velocity is zero
Impulse [tex]J=P_f-P_i[/tex]
[tex]J=0-(-0.25)[/tex]
[tex]J=0.25 kg-m/s[/tex]
A crate is lifted vertically 1.5 m and then heldat rest. The crate has weight 100 N (i.e., it issupported by an upward force of 100 N).How much work was done in lifting thecrate from the ground to its final position?1. More than 150 J2. 150 J3. A bit less than 150 J4. No work was done.5. None of these
Answer:
option 1
Explanation:
given,
Weight of crate = 100 N
Crate is lifted up to height = 1.5 m
Work done =?
work = Force x distance
work = 100 N x 1.5 m
work = 150 J
However, work would be more than the 150 J to lift the crate slightly higher to its final position.
The correct answer is option 1
Water flowing through a cylindrical pipe suddenly comes to a section of the pipe where the diameter decreases to 86% of its previous value. If the speed of the water in the larger section of the pipe was 32 m/s what is its speed in this smaller section if the water behaves like an ideal incompressible fluid?
Answer:
The speed in the smaller section is [tex]43.2\,\frac{m}{s}[/tex]
Explanation:
Assuming all the parts of the pipe are at the same height, we can use continuity equation for incompressible fluids:
[tex] \Delta Q=0 [/tex] (1)
With Q the flux of water that is [tex] Av[/tex] with A the cross section area and v the velocity, so by (1):
[tex] A_{2}v_{2}-A_{1}v_{1}=0 [/tex]
subscript 2 is for the smaller section and 1 for the larger section, solving for [tex] v_{2} [/tex]:
[tex]v_{2}=\frac{A_{1}v_{1}}{A_{2}} [/tex] (2)
The cross section areas of the pipe are:
[tex] A_{1}=\frac{\pi}{4}d_{1}^{2} [/tex]
[tex] A_{2}=\frac{\pi}{4}d_{2}^{2} [/tex]
but the problem states that the diameter decreases 86% so [tex] d_{2}=0.86d_{1} [/tex], using this on (2):
[tex] v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\approx1.35v_{1} [/tex]
[tex]v_{2}\approx(1.35)(32)\approx43.2\,\frac{m}{s} [/tex]
108J of work was done on a closed sysem. During this phase of the experiment, 79J of heat was added to the system.What was the total change in the internal energy of the system.
The change in internal energy of the system is +187 J
Explanation:
According to the first law of thermodynamics, the change in internal energy of a system is given by the equation:
[tex]\Delta U = Q + W[/tex]
where
[tex]\Delta U[/tex] is the change in internal energy
Q is the heat absorbed by the system
W is the work done on the system
For the system in this problem, we have
W = +108 J is the work done on it
Q = +79 J is the hear added to it
So, the change in internal energy is
[tex]\Delta U = 108 + 79 = +187 J[/tex]
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A father racing his son has half the kinetic energy of the son, whohas three-fifths the mass of the father. The father speeds up by2.5 m/s and then has the same kinetic energy as the son.a) What is the original speed of the father?b) What is the original speed of the son?
Answer:
a) 6.04 m/s
b) 11.02 m/s
Explanation:
a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s
[tex]E_2 = 2E_1[/tex]
[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]
[tex]V^2 + 5V + 6.25 = 2V^2[/tex]
[tex]V^2 - 5V - 6.25 = 0[/tex]
[tex]V \approx 6.04m/s[/tex]
b) The son kinetic energy initially is:
[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]
We can solve for the son speed by the following formula
[tex]E_s = \frac{mv^2}{2}[/tex]
[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]
[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]
Beth exerts 14 Newton’s of force to propel a 4.5 kilogram bowling ball down the lane. Describe how the ball will travel.
The ball will accelerate at a rate of [tex]3.11 m/s^2[/tex]
Explanation:
We can describe the motion of the ball by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass
a is the acceleration
In this problem,
F = 14 N is the force exerted on the ball
m = 4.5 kg is the mass of the ball
Solving the equation, we find its acceleration:
[tex]a=\frac{F}{m}=\frac{14}{4.5}=3.11 m/s^2[/tex]
So, the ball will accelerate at a rate of [tex]3.11 m/s^2[/tex].
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Assume the speed of light to be 299 792 458 m/s. If the frequency of an electromagnetic wave is 80,000 GHz (GHz = gigahertz = 109 Hz), what is the wavelength of that radiation? Express your answer in micrometres (μm)
Answer:
3.747 μm
Explanation:
To answer this question, the fundamental wave equation will be used. Light is an electromagnetic wave so we will use the speed of light for this electromagnetic wave.
v = fλ
299 792 458 m/s = 80,000 *10^9 * λ
λ = 3.747 *10^-6 = 3.747 μm
In order to sail through the frozen Arctic Ocean, the most powerful icebreaker ever built was constructed in the former Soviet Union. At the heart of the ship’s power plant is a nuclear reactor with a power output of 5.60* 10^7 W. How long will it take for this power plant to do 5.35* 10^10 J of work?
Answer:
955.36 seconds ≈ 16 minutes
Explanation:
Power(P) is the rate of doing work(W)
That is, P = W/t, where t is the time.
multipying both sides with 't' and dividing with 'P', we get: t=W/P
Here, W = 5.35 x 10^10 J and P = 5.6 x 10^7 W ( 1 W = 1 J/s).
Therefore , on dividing W with P, we get 955.36 seconds.
The cheetah is one of the fastest-accelerating animals, because it can go from rest to 19.6 m/s (about 44 mi/h) in 2.9 s. If its mass is 108 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
(a) watts(b) horsepower.
Answer:
a)P =14288.4 W
b)P = 19.16 horsepower
Explanation:
Given that
m= 108 kg
Initial velocity ,u= 0 m/s
Final velocity ,v= 19.6 m/s
t= 2.9 s
Lets take acceleration of Cheetah is a m/s²
We know that
v= u + a t
19.6 = 0 + a x 2.9
a= 6.75 m/s²
Now force F
F= m a
F= 108 x 6.75 N
F= 729 N
Now the power P
P = F.v
P = 729 x 19.6 W
P =14288.4 W
We know that
1 W= 0.0013 horsepower
P = 19.16 horsepower
P =14288.4 W
A batter hits a 0.140-kg baseball that was approaching him at 19.5 m/s and, as a result, the ball leaves the bat at 44.8 m/s in the reverse of its original direction. The ball remains in contact with the bat for 1.7 ms. What is the magnitude of the average force exerted by the bat on the ball?
Answer:
5295.3 N
Explanation:
According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force
mv + P = mV
P = mV - mv = m(V - v)
Since the velocity of the ball before and after is in opposite direction, one of them is negative
P = 0.14(44.8 - (-19.5)) = 9 kg m/s
Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is
F = P/t = 9/0.0017 = 5295.3 N
Classify the given types of matter as either baryonic (ordinary matter that contains protons and neutrons) or as nonbaryonic ("extraordinary" matter that consists of more exotic subatomic particles)
(Select B - Baryonic, N - Nonbaryonic. If the first is B and the rest N, enter
BNNNNN).
a. matter in our bodies
b. dark matter consisting of weakly interacting subatomic particles
c. dark matter consisting of Jupiter-sized planets in galactic halos
d. matter in brown dwarfs
e. matter that probably makes up the majority of dark matter in the universe
Answer:
a) B
b) N
c) B
d) B
e) N
Explanation:
a) The matter in our bodies is the regular matter, basically, we are made of carbon molecules and water. So it involves baryonic matter.
b) In this case, the weakly interacting subatomic particles know as (WIMPs), is the primary candidate for dark matter and this kind of particle has not yet been discovered. We are talking about the nonbaryonic matter.
c) Planets as a Jupiter are made of baryonic matter, in the specific case of Jupiter, it is approximately 75% hydrogen and 24% helium by mass and they are baryonic matter.
d) By definition, brown dwarfs are objects which have a size between a giant gaseous planet like Jupiter and a smaller star, so using the definitions above they are made of baryonic matter.
e) The majority of dark matter is made of non-baryonic matter.
Sound wave A delivers 2J of energy in 2s. Sound wave length B delivers 10J of energy in 5s. Sound wave C delivers 2mJ of energy in 1ms. Rank in order, from largest to smallest, the sound powers of Pa, Pb, Pc of these three waves.Explain. What equation would you use to determine this?
Answer:
[tex]P_c=P_b>P_a[/tex]
Explanation:
E = Energy
T = Time
Power is given by the equation
[tex]P=\frac{E}{T}[/tex]
For first case
[tex]P_a=\frac{2}{2}\\\Rightarrow P_a=1\ W[/tex]
For second case
[tex]P_b=\frac{10}{5}\\\Rightarrow P_b=2\ J[/tex]
For third case
[tex]P_c=\frac{2\times 10^{-3}}{1\times 10^{-3}}\\\Rightarrow P_b=2\ J[/tex]
The rank of power would be [tex]P_c=P_b>P_a[/tex]
A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change in both temperature and pressure. find density of the gas
Answer:
[tex]\rho = 0.50 g/L[/tex]
Explanation:
As we know that
PV = nRT
here we have
[tex]P = 1.0 atm[/tex]
[tex]P = 1.013 \times 10^5 Pa[/tex]
so we have
[tex]V = 1.4 \times 10^{-3} m^3[/tex]
T = 290 K
now we have
[tex](1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)[/tex]
[tex]n = 0.06 [/tex]
now the mass of gas is given as
[tex]m = n M[/tex]
[tex]m = (0.06)(28)[/tex]
[tex]m = 1.65 g[/tex]
now density of gas when its volume is increased to 3.3 L
so we will have
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{1.65 g}{3.3 L}[/tex]
[tex]\rho = 0.50 g/L[/tex]
Final answer:
The density of a gas can be calculated using the ideal gas law, which states that PV = nRT. To find the density, we can rearrange the equation and substitute the given values.
Explanation:
The density of a gas can be calculated using the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. To find the density, we can rearrange the equation as follows:
Density = mass/volume = (molar mass * n) / V
Using the given information, we can substitute the values into the equation and solve for the density.
Linear Thermal Expansion (in one dimension)
1) The change in length ΔL is proportional to the original length L, and the change in temperature ΔT : ΔL = αLΔT, where ΔL is the change in length , and α is the coefficient of linear expansion.
a) The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest (–15ºC). The coefficient of linear expansion, α , for steel is 12×10−6 /ºC. When the temperatures rises to 25 °C, what is its change in length in meters?
2) The change in volume ΔV is very nearly ΔV ≈ 3αVΔT . This equation is usually written as ΔV = βVΔT, where β is the coefficient of volume expansion and β ≈ 3α . V is the original volume. ΔT is the change in temperature. Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gasoline, and both the tank and the gasoline have a temperature of 15.0ºC . The coefficients of volume expansion, for gasoline is βgas = 950×10−6 /ºC , for the steel tank is βsteel = 35×10−6 /ºC .
a) What is the change in volume (in liters) of the gasoline when the temperature rises to 25 °C in L?
b) What is the change in volume (in liters) of the tank when the temperature rises to 25 °C in L?
c) How much gasoline would be spilled in L?
Answer:
1) [tex]\Delta L= 0.612\ m[/tex]
2) a. [tex]\Delta V_G=0.57\ L[/tex]
b. [tex]\Delta V_S=0.021\ L[/tex]
c. [tex]V_0=0.549\ L[/tex]
Explanation:
1)
given initial length, [tex]L=1275\ m[/tex]initial temperature, [tex]T_i=-15^{\circ}C[/tex]final temperature, [tex]T_f=25^{\circ}C[/tex]coefficient of linear expansion, [tex]\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}[/tex]∴Change in temperature:
[tex]\Delta T=T_f-T_i[/tex]
[tex]\Delta T=25-(-15)[/tex]
[tex]\Delta T=40^{\circ}C[/tex]We have the equation for change in length as:
[tex]\Delta L= L.\alpha. \Delta T[/tex]
[tex]\Delta L= 1275\times 12\times 10^{-6}\times 40[/tex]
[tex]\Delta L= 0.612\ m[/tex]
2)
Given relation:
[tex]\Delta V=V.\beta.\Delta T[/tex]
where:
[tex]\Delta V[/tex]= change in volume
V= initial volume
[tex]\Delta T[/tex]=change in temperature
initial volume of tank, [tex]V_{Si}=60\ L[/tex]initial volume of gasoline, [tex]V_{Gi}=60\ L[/tex]initial temperature of steel tank, [tex]T_{Si}=15^{\circ}C[/tex]initial temperature of gasoline, [tex]T_{Gi}=15^{\circ}C[/tex]coefficients of volumetric expansion for gasoline, [tex]\beta_G=950\times 10^{-6}\ ^{\circ}C[/tex]coefficients of volumetric expansion for gasoline, [tex]\beta_S=35\times 10^{-6}\ ^{\circ}C[/tex]a)
final temperature of gasoline, [tex]T_{Gf}=25^{\circ}C[/tex]
∴Change in temperature of gasoline,
[tex]\Delta T_G=T_{Gf}-T_{Gi}[/tex]
[tex]\Delta T_G=25-15[/tex]
[tex]\Delta T_G=10^{\circ}C[/tex]
Now,
[tex]\Delta V_G= V_G.\beta_G.\Delta T_G[/tex]
[tex]\Delta V_G=60\times 950\times 10^{-6}\times 10[/tex]
[tex]\Delta V_G=0.57\ L[/tex]
b)
final temperature of tank, [tex]T_{Sf}=25^{\circ}C[/tex]
∴Change in temperature of tank,
[tex]\Delta T_S=T_{Sf}-T_{Si}[/tex]
[tex]\Delta T_S=25-15[/tex]
[tex]\Delta T_S=10^{\circ}C[/tex]
Now,
[tex]\Delta V_S= V_S.\beta_S.\Delta T_S[/tex]
[tex]\Delta V_S=60\times 35\times 10^{-6}\times 10[/tex]
[tex]\Delta V_S=0.021\ L[/tex]
c)
Quantity of gasoline spilled after the given temperature change:
[tex]V_0=\Delta V_G-\Delta V_S[/tex]
[tex]V_0=0.57-0.021[/tex]
[tex]V_0=0.549\ L[/tex]
Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2. Gun #1 weighs three times as much as gun #2. The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:
a) 2:3
b) 3:2
c) 2:1
d) 3:1
e) 6:1
Answer:
option C
Explanation:
Let mass of the bullet be m and velocity be v
mass of gun be M and bullet be V
now,
using conservation of momentum for gun 1
(M+m) V' = 2 mv + 3 MV
V' = 0
3 M V = - 2 mv
momentum of gun 1 =- 2 mv---------(1)
now for gun 2
(M+m) V' = mv + MV
V' = 0
M V = - mv
momentum of gun 1 = -mv-----------(2)
dividing equation (1) by (2)
[tex]\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}[/tex]
[tex]\dfrac{P_m1}{P_m2} = \dfrac{2}{1}[/tex]
the correct answer is option C
The correct option is Option C (2:1).The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is 2:1. This conclusion follows from the principle of conservation of momentum. Thus, the correct answer is option c) 2:1.
When the bullets are fired, each gun experiences a momentum change due to the bullet's ejection. By the conservation of momentum, the momentum imparted to each gun should equal the momentum of its corresponding bullet in magnitude but in the opposite direction.Define the mass of the bullet from gun #2 as m. The bullet from gun #1 then has a mass of 2m since it is twice as heavy. Denote the velocity of both bullets as v:
The momentum of the bullet from gun #2 is p2 = m * v.The momentum of the bullet from gun #1 is p1 = 2m * v.So, the momentum imparted to gun #1 is 2m * v, and for gun #2 it is m * v.
To find the ratio of the momentum imparted to gun #1 to that imparted to gun #2, we calculate:
Ratio = (momentum imparted to gun #1) / (momentum imparted to gun #2) = (2m * v) / (m * v) = 2/1 = 2:1Therefore, the correct answer is option c) 2:1
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats per second. The tuner increases the tension on the as-yet untuned string, and now when they are played together the beat frequency is
1.0s−1.
(a) What was the original frequency of the untuned string?
(b) By what percentage did the tuner increase the tension on that string?
Final answer:
To find the two possible frequencies of the untuned piano string, we can use the formula for beat frequency. From the given information, the original frequency of the untuned string can be either 266.0 Hz or 262.0 Hz. To find the percentage increase in tension on the untuned string, we can use the formula for calculating percentage increase.
Explanation:
To find the two possible frequencies of the untuned piano string, we can use the formula for beat frequency:
Beat frequency = |Frequency of the first string - Frequency of the second string|
In this case, the beat frequency is given as 2.00 s. The frequency of the first string is 264.0 Hz. Let's assume the frequency of the second string is x Hz.
So, we can set up the equation:
2.00 = |264.0 - x|
Solving for x, we get two possible frequencies: 266.0 Hz and 262.0 Hz.
To find the original frequency of the untuned string, we can use the formula:
Original frequency = Frequency of the first string ± Beat frequency
For positive beat frequencies, the original frequency would be:
Original frequency = 264.0 + 2.00 = 266.0 Hz
For negative beat frequencies, the original frequency would be:
Original frequency = 264.0 - 2.00 = 262.0 Hz
To find the percentage increase in tension on the untuned string, we can use the formula:
Percentage increase = (Change in tension / Original tension) x 100
Since the tension on the first string is unchanged (as it is the tuned string), the change in tension on the untuned string is equal to the change in frequency. Assuming the original frequency of the untuned string is 262.0 Hz:
Change in tension = |Original frequency - New frequency|
Change in tension = |262.0 - 264.0| = 2.0 Hz
Therefore, the percentage increase in tension on the untuned string is:
(2.0 / 262.0) x 100 = 0.763%
A 12 inch telescope has an angular resolution how many times smaller when compared to a 4 inch telescope?
Answer:
θ₂ = 3 θ₁
Explanation:
given,
telescope of lens diameter = 12 inch
another telescope of lens diameter = 4 inch
comparison of resolution power.
Using the formula of resolution
[tex]\theta = \dfrac{1.22 \lambda}{D}[/tex]
for diameter = 12 inch
[tex]\theta_1 = \dfrac{1.22 \lambda}{D_1}[/tex].....(1)
for diameter = 4 inch
[tex]\theta_2 = \dfrac{1.22 \lambda}{D_2}[/tex].......(2)
dividing equation (2) from (1)
[tex]\dfrac{\theta_2}{\theta_1} = \dfrac{D_1}{D_2}[/tex]
now,
[tex]\dfrac{\theta_2}{\theta_1} = \dfrac{12}{4}[/tex]
[tex]\dfrac{\theta_2}{\theta_1} =3[/tex]
θ₂ = 3 θ₁
hence, we can say that resolution of telescope of 12 inch is 3 time smaller than the resolution of 4 inch telescope.
The end point of a spring vibrates with a period of 2.1 seconds when a mass m is attached to it. When this mass is increased by 6.810×101 kg, the period is found to be 3.4 seconds. Find the value of m.
Answer:
Mass attached to the spring is 41.95 kg
Explanation:
We have given time period of the spring T = 2.1 sec
Let the mass attached is m
And spring constant is k
We know that time period is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1
Now if the mass is increased by 68.10 kg then time period become 3.4 sec
So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2
Now dividing eqn 1 by eqn 2
[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]
[tex]0.381=\frac{m}{m+68.10}[/tex]
[tex]m=41.95 kg[/tex]
So mass attached to the spring is 41.95 kg
Final answer:
To find the value of mass m, use the formula for the period of a mass-spring system.
Explanation:
In order to find the value of mass m, we can use the formula for the period of a mass-spring system:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
For the initial system with period 2.1 seconds, we have:
2.1 = 2π√(m/k)
For the system with mass increased by 6.810×10^1 kg and period 3.4 seconds, we have:
3.4 = 2π√((m + 6.810×10^1)/k)
Using these two equations, we can solve for the value of m.
The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a height 6450 km above Earth’s surface? Answer in units of N. What is the weight 33700 km above Earth’s surface? Answer in units of N.
Final answer:
The weight of a spacecraft at 6450 km above Earth's surface is 1767.5 N and at 33700 km above Earth's surface is 182.35 N, calculated using Newton's law of universal gravitation and considering the increased distance from the Earth's center.
Explanation:
The weight of a spacecraft can be calculated using Newton's law of universal gravitation which states that every mass attracts every other mass with a force that is directly proportional to the product of their masses and inversely proportional to the distance squared between their centers, F = G * (m₁ * m₂) / r², where G is the gravitational constant, m₁ and m₂ are the masses involved, and r is the distance between the centers of the two masses. To find the weight of the spacecraft at a certain height, we need to use the spacecraft's mass and the new distance from the Earth's center, which includes both the Earth's radius and the altitude above the surface.
To answer the first part of the question, we calculate the weight at 6450 km above Earth's surface. Since the radius of the Earth is also 6450 km, the distance from the center of the Earth to the spacecraft is now 2 * 6450 km. Applying the law of gravitation, the gravitational force, and hence the weight, will be (6450 km / 2 * 6450 km)² = 1/4 of the original weight, which is 7070 N/4 = 1767.5 N.
For the second part of the question, at a height of 33700 km above the Earth's surface, the distance from the center is 6450 km + 33700 km = 40150 km. Repeating the calculation, the weight at this height will be (6450 km / 40150 km)² times the original weight, giving us a reduced weight of (7070 N * (1/6.23)²) ≈ 182.35 N.
A sound can be _______ or ________.
A. Quiet; loud
B. Quiet; soft
C. Quiet; heavy
D. Loud; soft
Answer:
a
Explanation:
a sound cannot be soft
Answer:
Quite or loud
Explanation:
Models that explain the formation of the solar system through a series of gradual steps are considered evolutionary theories. What are models that explain the formation of certain objects in the solar system through other means considered?
Answer:
Catastrophic theories
Explanation:
The theory of catastrophe is a collection of methods used to analyze and describe the ways in which a system can experience sudden significant behavioral changes when one or more of the variables that govern it are continuously modified.
Georges Louis de Buffon suggested in 1745 the first destructive theory — that a comet pulled material from the Sun to form the planets.
Therefore the answer is -
Catastrophic theories
Describe what happens, at a microscopic level, when an object is charged by rubbing. For instance, what happens when a plastic pipe is rubbed with a cloth? Describe the specific case where the rod becomes negatively charged
Answer:
Explanation:
The static charges are generated due to excess or deficiency of electrons, because these are the smallest quanta of charge available at the molecular level which can get transferred with minimal energy requirement.
These charges are usually generated by friction between the two surfaces leading to the transfer of electron from one to another.
When a plastic pipe is rubbed with a cloth then due to friction the surface of the cloth loses electron which gets stuck at the surface of the pipe making it negatively charged.
A ledge on a building is 23 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up over a pulley and straight down to a 3.0-kg can of nails on the ground. If the can of paint is accidentally knocked off the ledge, what time interval does a carpenter have to catch the can of paint before it smashes on the ground?
Answer:
The time can catch before it smashes on the ground is [tex]t=5.73 s[/tex]
Explanation:
Using the force equation
[tex]F=m*a[/tex]
[tex]F_{net}=m*a[/tex]
So replacing and solving to find the acceleration
[tex]a = (m_1*g-m_2*g) / m_1+m_2[/tex]
Finding the factor
[tex]a = g *( m_1-m_2)/m_1+m_2[/tex]
[tex]a=9.8m/s^2 *( 4.0 kg- 3.0 kg) / (4.0 + 3.0) kg[/tex]
[tex]a=1.4 m/s^2[/tex]
Now replacing in Newtons law to find the time before can catch so:
[tex]d= \frac{1}{2}*a*t^2[/tex]
[tex]t=\sqrt{\frac{2*d}{a}}=\sqrt{\frac{2* 23m}{1.4 m/s^2}}[/tex]
[tex]t=5.73 s[/tex]
A human being can be electrocuted if a current as small as 48 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding. If his resistance is 2100 Ω, what might the fatal voltage in volts be?
Answer:
V = 100.8 V
Explanation:
given,
hum being can be electrocuted with current = 48 mA = 0.048 A
Resistance of the man = 2100 Ω
Fatal voltage = ?
we know,
V = I R
V is the fatal voltage in Volts
R is the resistance provided by the human body
I is current
V = I R
V = 0.048 x 2100
V = 100.8 V
the voltage which can be considered as fatal is equal to V = 100.8 V
You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a
Answer:
Sa
Explanation:
Spiral Galaxies -
It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .
According to Hubble , these galaxies are classified as Sa , Sb , Sc .
Where ,
Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .
Sb - have a lesser bulge and the spiral is looser .
Sc - It has very weak bulge with the open spiral structure .
Hence , from the question ,
The given information is about the Sa .
The melting of glaciers and the polar ice caps poses a devastating threat of potential flooding for low-lying areas. Scientists predict that a sea level rise of ________ would inundate areas such as the delta regions of africa and asia
Answer:
4-6 millimeters
Explanation:
Global warming is causing devastating consequences for the planet such as rising sea levels and temperature in the oceans.
The melting of glaciers is one of the main causes of sea level rise.
Undoubtedly, the most affected and vulnerable areas correspond to areas of Asia, Africa and South America. Specifically, the study highlights that in six Asian countries such as China, Bangladesh, India, Vietnam, Indonesia and Thailand, there are approximately 237 million people who will suffer these floods if defense mechanisms are not activated.
A 0.095-kg aluminium sphere is dropped from the roof of a 55-m-high building. The specific heat of aluminium is 900 J/kg⋅C∘ .
If 65 % of the thermal energy produced when it hits the ground is absorbed by the sphere, what is its temperature increase?
Answer:
Increase in temperature will be [tex]0.389^{\circ}C[/tex]
Explanation:
We have given mass of the aluminium m = 0.095 kg
Height h = 55 m
Specific heat of aluminium c = 900 J/kg°C
We know that potential energy is given as
[tex]PE=mgh=0.095\times 9.8\times 55=51.205[/tex]
Now 65 % of potential energy [tex]=\frac{51.205\times 65}{100}=33.28[/tex]
Now this energy is used to increase the temperature
So [tex]mc\Delta T=33.28[/tex]
[tex]0.095\times 900\times \Delta T=33.28[/tex]
[tex]0.095\times 900\times \Delta T=33=0.389^{\circ}C[/tex]
What parts of the nucleotide make up the backbone of the dna molecule
The backbone of the DNA molecule is formed by alternating sugar and phosphate groups. The nitrogenous bases are located in the interior of the molecule.
Explanation:The backbone of the DNA molecule is made up of the alternating sugar and phosphate groups. The sugar and phosphate groups are bonded by covalent bonds, and they line up on the outside of each strand. The nitrogenous bases, which include adenine (A), thymine (T), cytosine (C), and guanine (G), are stacked in the interior of the DNA molecule.
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The backbone of a DNA molecule is formed from alternating sugar and phosphate groups of nucleotides. The nitrogenous bases, which are not part of the backbone, protrude from it and are involved in the formation of the double helix structure.
Explanation:The backbone of the DNA molecule consists of alternating sugar and phosphate groups. A nucleotide, which is the building block of DNA, consists of three components: a nitrogenous base, a pentose sugar, and a phosphate group. The backbone is formed by the bonding of the phosphate group of one nucleotide to the sugar of the next nucleotide, creating a chain of sugar-phosphate bonds.
For instance, the phosphate group is attached to the 5' carbon of one nucleotide and the 3' carbon of the next nucleotide. In this sense, the DNA molecule can be visualized as a twisted ladder where the backbone represents the rails of the ladder and the nitrogenous bases represent the steps.
It's important to note that the nitrogenous bases are not part of the backbone; They stick out from the backbone and are involved in hydrogen bonding with the nitrogenous bases of the complementary DNA strand, resulting in a double helix structure. The two strands of the DNA run in opposite directions making them antiparallel.
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An apparatus like the one Cavendish used to find G has large lead balls that are 8.4 kg in mass and small ones that are 0.061 kg. The center of a large ball is separated by 0.057 m from the center of a small ball. Find the magnitude of the gravitational force between the masses if the value of the universal gravitational constant is 6.67259 × 10−11 Nm2/kg2
Answer:
The gravitational force is [tex]1.05\times10^{-8}\ N[/tex]
Explanation:
Given that,
Mass of large ball = 8.4 kg
Mass of small ball = 0.061 kg
Separation = 0.057 m
Gravitational constant [tex]G= 6.67\times10^{-11}\ Nm^2/kg^2[/tex]
We need to calculate the gravitational force
Using formula of gravitational force
[tex]F= \dfrac{Gm_{1}m_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{6.67259\times10^{-11}\times8.4\times0.061}{(0.057)^2}[/tex]
[tex]F=1.05\times10^{-8}\ N[/tex]
Hence, The gravitational force is [tex]1.05\times10^{-8}\ N[/tex]
To find the gravitational force between two masses, you use the formula derived from Newton's law of universal gravitation, F = G × (m1 × m2) / r², and with the provided values, the force is calculated to be approximately 1.19 × 10⁻¹° Newtons.
Explanation:The student has asked about the gravitational force between two masses using the apparatus similar to the one used in the Cavendish experiment. To calculate the magnitude of the gravitational force between the large lead balls (8.4 kg each) and the small balls (0.061 kg each), separated by a distance of 0.057 m, and using the universal gravitational constant (G = 6.67259 × 10⁻¹¹ Nm²/kg²), the following formula derived from Newton's law of universal gravitation is used:
F = G × (m1 × m2) / r²
Substituting the given values:
F = (6.67259 × 10⁻¹¹) × (8.4 × 0.061) / (0.057²)
After performing the calculation, we find that the gravitational force F is approximately 1.19 × 10⁻¹° Newtons. This force is a direct application of the universal law of gravitation, indicating that two masses will always exert a gravitational pull on each other, no matter how small.