Answer:
P_2 = 62.69 psi
Explanation:
given,
P₁ = 70 psia T₁ = 55° F = (55 + 459.67) R
P₂ = ? T₂ = 115° F = (115 + 459.67) R
we know,
p = ρ RT
ρ is the density which is constant
R is also constant
now,
[tex]\dfrac{P_2}{P_1} =\dfrac{T_2}{T_1}[/tex]
[tex]\dfrac{P_2}{70} =\dfrac{55+459.67}{115+459.67}[/tex]
P_2 = 62.69 psi
Hence, the increase in Pressure is equal to P_2 = 62.69 psi
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)
Answer:
532.0725 m
102.17270893 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = g
H = Height of cliff
Distance traveled in 3 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]
Distance traveled by sound = 2H-44.145 m
[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]
The height of the cliff is 532.0725 m
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]
Her speed just before she hits the ground is 102.17270893 m/s
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the following: (a) The velocity v and elevation y of the ball above the ground at any time t. (b) The highest elevation reached by the ball and its corresponding time t. (c) The time when the ball will hit the ground and the impact velocity.
Answer:
Explanation:
Given
Initial velocity of ball [tex]u=10\ m/s[/tex]
height of window [tex]h=20\ m[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
t=time
a=acceleration
As ball is already is at a height of 20 m so
[tex]Y=ut+\frac{1}{2}at^2+20[/tex]
[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]
[tex]Y=-4.9t^2+10t+20[/tex]
(b)highest point is obtained at v=0
[tex]v^2-u^2=2as[/tex]
where
v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](0)-10^2=2\times (-9.8)\times s[/tex]
[tex]s=\frac{100}{19.6}[/tex]
[tex]s=5.102\ m[/tex]
Highest Point will be [tex]s+20=25.102\ m[/tex]
(c)Time taken when the ball hit the ground i.e. at Y=0
[tex]-4.9t^2+10t+20=0[/tex]
[tex]t=3.28\ s[/tex]
impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]
[tex]v=22.181\ m/s[/tex]
(a) The equation be "Y = -4.9t² + 10t + 20".
(b) The highest point be "25.102 m".
(c) The impact velocity be "22.181 m/s"
Equation of motionAccording to the question,
Ball's initial velocity, u = 10 m/s
Window's height, h = 20 m
(a) By using equation of motion,
Y = ut + [tex]\frac{1}{2}[/tex]at²
By substituting the values,
= ut + [tex]\frac{1}{2}[/tex]at² + 20
= 10 × t + 0.5 × (9.8)t² + 20
= -4.9t² + 10t + 20
(b) We know that,
→ v² - u² = 2as
here, Final velocity, v = 0
0 - (10)² = 2 × (-9.8) × s
s = [tex]\frac{100}{19.6}[/tex]
= 5.102 m
(c) Time taken will be:
→ -4.9t² + 10t + 20 = 0
t = 3.28 s
hence,
The impact velocity,
v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]
= 22.181 m/s
Thus the above response is correct.
Find out more information about velocity here:
https://brainly.com/question/6504879
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\times 10^{-5}~\text{N}\cdot\text{m}^2/\text{C}3×10 −5 N⋅m 2 /C when the area is parallel to the sheet of charge. Find the charge density on the sheet.
Answer:
Explanation:
Given
side of square shape [tex]a=5\ cm[/tex]
Electric flux [tex]\phi =3\times 10^{-5}\ N.m^2/C[/tex]
Permittivity of free space [tex]\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}[/tex]
Flux is given by
[tex]\phi =EA\cos \theta [/tex]
where E=electric field strength
A=area
[tex]\theta [/tex]=Angle between Electric field and area vector
[tex]E=\frac{\phi }{A\cos (0)}[/tex]
[tex]E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}[/tex]
[tex]E=0.012\ N/C[/tex]
and Electric field by a uniformly charged sheet is given by
[tex]E=\frac{\sigma }{2\epsilon_0}[/tex]
where [tex]\sigma[/tex]=charge density
[tex]=\frac{\sigma }{\epsilon_0}[/tex]
[tex]\sigma =0.012\times 8.85\times 10^{-12}[/tex]
[tex]\sigma =2.12\times 10^{-13}\ C/m^2[/tex]
How many kWh of energy does a 550-W toaster use in the morning if it is in operation for a total of 5.0 min? At a cost of 9.0 cents/k Wh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week.
Answer:
0.0458 kWh
6.5736 cents
Explanation:
The formula for electric energy is given as
E = Pt................. Equation 1
Where E = Electric energy, P = Electric power, t = time.
Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.
Substituting into equation 1
E = 550(0.083)
E = 45.83 Wh
E = (45.83/1000) kWh
E = 0.0458 kWh.
Hence the kWh = 0.0458 kWh.
If the makes a toast four morning per week, and the are Four weeks in a month.
Total number days he makes toast in a month = 4×4 = 16 days.
t = 16×0.083 h = 1.328 h.
Total energy used in a month = 550(1.328)
E = 730.4 Wh
E = 0.7304 kWh.
If the cost of energy is 9.0 cents per kWh,
Then for 0.7304 kWh the cost will be 9.0(0.7304) = 6.5736 cents.
Hence this would add 6.5736 cents to his monthly electric bill
A 550-W toaster in operation for 5.0 minutes uses 2.75 kWh of energy. If you make toast four mornings per week, it would add an estimated cost of $4.30 to your monthly electric energy bill.
Explanation:To calculate the energy used by the toaster, we can use the formula E = Pt, where P is the power and t is the time. In this case, the power of the toaster is 550 watts and the time it is in operation is 5.0 minutes. Plugging these values into the formula, we get E = (550 W)(5.0 min) = 2750 W.min. To convert this to kilowatt-hours (kWh), we need to divide by 1000, so the energy used by the toaster is 2.75 kWh.
To estimate how much this would add to your monthly electric energy bill, we need to know how many times you use the toaster in a month. If you use it four mornings per week, that would be 4 days x 52 weeks / 12 months = 17.33 days per month. Multiplying the energy used by the toaster (2.75 kWh) by the number of days in a month (17.33), we get an estimate of 47.75 kWh per month. Finally, to find the cost, we multiply the energy (47.75 kWh) by the cost per kilowatt-hour (9.0 cents/kWh) and convert it to dollars, giving us an estimated cost of $4.30 per month.
Learn more about Calculating energy consumption of a toaster here:https://brainly.com/question/15569623
#SPJ3
You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located in the plane of the spacecraft’s orbit. You find that the spacecraft’s radio signal varies periodically in wavelength between 2.99964 m and 3.00036 m. Assuming that the radio is broadcasting at a constant wavelength, what is the mass of the planet?
To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:
[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]
Here,
[tex]v_{o} =[/tex] Orbital Velocity
[tex]\lambda_{max} =[/tex] Maximal Wavelength
[tex]\bar{\lambda}} =[/tex] Average Wavelength
c = Speed of light
Replacing with our values we have that,
[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]
Note that the average signal is 3.000000m
[tex]v_o = 36 km/s[/tex]
Now using the definition about centripetal acceleration we have,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Orbit Velocity
r = Radius of Orbit
Replacing with our values,
[tex]a = \frac{(36km/s)^2}{100000km}[/tex]
[tex]a= 0.01296km/s^2[/tex]
[tex]a = 12.96m/s^2[/tex]
Applying Newton's equation for acceleration due to gravity,
[tex]a =\frac{GM}{r^2}[/tex]
Here,
G = Universal gravitational constant
M = Mass of the planet
r = Orbit
The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,
[tex]M = \frac{ar^2}{G}[/tex]
[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]
[tex]M = 1.943028*10^{27}kg[/tex]
Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]
The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken by Pluto to orbit the Sun in Earth days.
Answer:
Explanation:
Given
Distance between Pluto and sun is 39.1 times more than the distance between earth and sun
According to Kepler's Law
[tex]T^2=kR^3[/tex]
where k=constant
T=time period
R=Radius of orbit
Suppose [tex]R_1[/tex] is the radius of orbit of earth and sun
so Distance between Pluto and sun is [tex]R_2=39.1\cdot R_1[/tex]
[tex]T_1[/tex] and [tex]T_2[/tex] is the time period corresponding to [tex]R_1[/tex] and R_2[/tex]
[tex](T_1)^2=k(R_1)^3---1[/tex]
[tex](T_2)^2=k(R_2)^3---2[/tex]
divide 1 and 2
[tex](\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3[/tex]
[tex]T_2^2=365^2\times 39.1^3[/tex]
[tex]T_2=89239.67\ Earth\ days[/tex]
Final answer:
Using Kepler's Third Law of Planetary Motion, we can calculate that it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
Explanation:
To calculate the time taken by Pluto to orbit the Sun, we use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Given that the distance between Pluto and the Sun is approximately 39.1 times more than the distance between the Earth and the Sun, we could simplify and use the approximation that Pluto's distance in astronomical units (AU) is roughly 40, as given in the reference information.
Pluto's orbital period (P) can be calculated using the average distance from the Sun (a) with the formula P² = a³. Here, we assume Earth's orbital period to be 1 Earth year and its distance as 1 AU by definition. Therefore, for Pluto:
a = 40 AU (Pluto's distance from the Sun) P² = 40³ P² = 64,000 P = √64,000 P ≈ 253 (Pluto's orbital period in Earth years)
To convert this period into Earth days, we multiply by the number of days in one Earth year (365.25 days, accounting for the leap year cycle):
P (in days) = 253 years × 365.25 days/year P (in days) ≈ 92,446.25 days
Hence, it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
A rock is thrown straight up into the air with an initial speed of 55 m/s at time t = 0. Ignore air resistance in this problem. At what times does it move with a speed of 36 m/s? Note: There are two answers to this problem.
Answer:
After 1.938 sec velocity of rock will be 36 m/sec
Explanation:
We have given initial velocity at which rock is thrown u = 55 m/sec
Final velocity v = 36 m/sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
According to first equation of motion we know that [tex]v=u+gt[/tex], here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time
So [tex]36=55-9.8t[/tex] ( Negative sign is due to rock is thrown upward )
So [tex]9.8t=19[/tex]
t = 1.938 sec
So after 1.938 sec velocity of rock will be 36 m/sec
If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? What forms of radiation might penetrate the clouds and reach the ground?
The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.
Through these waves, you can know that there is beyond the clouds.
Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.
Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.
It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.
Explanation:Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.
Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.
Learn more about radiation here:
https://brainly.com/question/4075566
#SPJ3
Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative electrode), where they are produced, toward the screen, which also serves as the anode (positive electrode), 25.0 cm away.At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest, and ignore relativistic effects?
Answer:
83816746.4254 m/s
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = [tex]2\times 10^4\ V[/tex]
The kinetic energy of the electron is
[tex]K=\dfrac{1}{2}mv^2[/tex]
Energy is given by
[tex]E=qV[/tex]
Balancing the energy
[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]
The velocity of the electrons is 83816746.4254 m/s
Calculate the molecular weight of a polyethylene molecule with n=750. Express your answer to three significant figures.
To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.
Explanation:To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).
Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol
Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.
Learn more about Calculating Molecular Weight of Polyethylene here:https://brainly.com/question/32556274
#SPJ3
In an experiment to measure the acceleration due to gravity g, two independent equally reliable measurements gave 9.67 m/s2 and 9.88 m/s2. Determine (i) the percent difference of the measurements (ii) the percent error of their mean. [Take the theoretical value of g to be 9.81 m/s
Answer:
i. +/- 1.43% and +/- 0.71% ii. +/- 0.33%
Explanation:
[tex]% Error = \frac{Error}{Measurement}* 100%\\[/tex]
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge’s arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically 12.5 g. (a) How fast must the astronaut’s head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is 2.00 m tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?
Answer:
32.9242311983 m/s
1.40506567727g
35.3405770759 rpm
Explanation:
v = Linear Velocity of the capsule
[tex]a_c[/tex] = Centripetal acceleration = [tex]12.5g=12.5\times 9.81[/tex]
r = Radius of the centrifuge = 8.84 m
l = Person's height = 2 m
Centripetal acceleration is given by
[tex]a_c=\frac{v^2}{r}\\\Rightarrow v=a_cr\\\Rightarrow v=\sqrt{12.5\times 9.81\times 8.84}[/tex]
The linear speed of the capsule is 32.9242311983 m/s
The radius would be
[tex]r=\sqrt{r^2+\dfrac{l^2}{4}}\\\Rightarrow r=\sqrt{8.84^2+\dfrac{2^2}{4}}\\\Rightarrow r=8.89638128679\ m[/tex]
The centripetal acceleration
[tex]a_{c2}=\dfrac{32.9242311983^2}{8.89638128679\times 9.81}g\\\Rightarrow a_{c2}=12.4207805891g[/tex]
Change in acceleration from Pythagoras law
[tex]a=\sqrt{a_{ch}^2-a_{c2}^2}\\\Rightarrow a=\sqrt{12.5^2g^2-12.4207805891^2g^2}\\\Rightarrow a=1.40506567727g[/tex]
The difference is 1.40506567727g
Velocity
[tex]v=\omega r\\\Rightarrow v=2\pi Nr\\\Rightarrow N=\dfrac{v}{2\pi r}\\\Rightarrow N=\dfrac{32.9242311983}{2\pi 8.89638128679}\\\Rightarrow N=0.589009617932\ rev/s[/tex]
[tex]N=0.589009617932\times 60=35.3405770759\ rpm[/tex]
The speed is 35.3405770759 rpm
A phonograph record has an initial angular speed of 37 rev/min. The record slows to 14 rev/min in 1.6 s. What is the record’s average angular acceleration during this time interval? Answer in units of rad/s 2
Answer:
Acceleration will be [tex]\alpha =-1.50rad/sec^2[/tex]
Explanation:
We have given initial angular velocity [tex]\omega _i=37rpm[/tex]
In radian/sec initial angular velocity will be [tex]\omega _i=\frac{2\times \pi 37}{60}=3.873rad/sec[/tex]
Angular velocity after 1.6 sec is 14 rpm
So final angular velocity [tex]\omega _f=\frac{2\times \pi\times 14}{60}=1.465rad/sec[/tex]
Time t = 1.6 sec
We have to find the angular angular acceleration
From first equation of motion we know that
[tex]\omega _f=\omega _+\alpha t[/tex]
[tex]1.465=3.873+\alpha \times 1.6[/tex]
[tex]\alpha =-1.50rad/sec^2[/tex] here negative sign indicates that motion is deaccelerative in nature
If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision?
Using the physics equation of motion and the given initial velocity, reaction time, and deceleration, one can determine whether a truck can stop in time to avoid a collision.
Explanation:The question focuses on stopping distance and acceleration required to avoid a collision, indicating its base in Physics. If we have a truck moving at a constant velocity and it brakes at a certain distance from an obstacle, the minimum acceleration needed to avoid a collision can be calculated using the equation of motion v^2 = u^2 + 2as. Here, 'v' is the final velocity (0 m/s as the truck needs to stop), 'u' is the initial velocity, 'a' is the acceleration, and 's' is the distance over which the truck needs to stop.
To determine if the truck will hit the child, we must account for the driver's reaction time as well. During this reaction time, the truck continues to travel at its initial speed. After the reaction time, the truck will begin decelerating until it comes to a stop. The total stopping distance is the distance covered during the reaction time plus the distance covered during deceleration. The latter can be found using the deceleration rate and the formula mentioned above.
For the given scenario of the truck with an initial velocity of 10 m/s, a braking distance of 50 m, reaction time of 0.5 seconds, and deceleration of -1.25 m/s^2, we can calculate whether or not the truck will be able to stop in time to avoid hitting the child.
An orange loses 1.2 kJ of heat as it cools per °C drop in its temperature. What is the amount of heat loss from the orange per °F drop in its temperature?
To solve this problem we will apply the conversion rate between Celcius and Fahrenheit degrees. We will use the direct relationship clearly and not the added degrees of scale conversion. We know from the statement that the orange loses heat at the rate of
[tex]Q = 1.2kJ/\°C[/tex]
We have the conversion to °F is given as
[tex]T (\°F) = 1.8T+32[/tex]
Calculate the amount of heat loss from orange per °F
[tex]Q = \frac{1.2}{1.8}[/tex]
[tex]Q = 0.667kJ/\°F[/tex]
Therefore the amount of heat loss from the orange per °F drop in its temperature is 0.667kJ/°F
The heat loss from an orange per °F drop is 0.67 kJ, calculated by taking 1.2 kJ per °C drop and dividing it by 1.8 to convert it to Fahrenheit,
Explanation:The heat loss from the orange per °F drop in its temperature can be found by converting 1.2 kJ lost per 1 °C drop in temperature to kJ lost per 1 °F drop. This can be achieved using the formula that 1 °C equals 1.8 °F.
Therefore, the heat loss per degree Fahrenheit will be less than the heat loss per degree Celsius. We calculate this as follows:
(1.2 kJ / °C) / 1.8 = 0.67 kJ per °F.
So for every degree Fahrenheit that the orange cools, it will lose 0.67 kilojoules of heat.
Learn more about Heat loss here:https://brainly.com/question/31857421
#SPJ3
A load consists of a 70-Ω resistor in parallel with a 90-μF capacitor. If the load is connected to a voltage source vs(t) = 160cos 2000t, find the average power delivered to the load.
Answer:
Power delivered by the source will be 182.912 watt
Explanation:
We have given a load is consist of a resistor of 70 ohm in parallel with [tex]90\mu F[/tex] capacitance
Voltage source is given [tex]v_s(t)=160cos(2000t)[/tex]
So maximum value of voltage source is 160 volt
So rms value [tex]v_{r}=\frac{v_m}{\sqrt{2}}=\frac{160}{1.414}=113.154volt[/tex]
We know that average power delivered by the source will be equal to average power absorbed by the resistor
So power absorbed by the resistor [tex]P=\frac{v_r^2}{R}=\frac{113.154^2}{70}=182.912watt[/tex]
So power delivered by the source will be 182.912 watt
A dripping water faucet steadily releases drops 1.0 s apart. As these drops fall, does the distance between them increase, decrease, or remain the same? Prove your answer.
Answer:
Distance between them increase
Explanation:
The position S of the water droplet can be determined using equation of motion
[tex]S=ut+\frac{1}{2} at^2[/tex]
where [tex]u[/tex] is the initial velocity which is zero here
[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity
the position of first drop after time [tex]t[/tex] is given by
[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]
the position of next drop at same time is
[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]
distance between them is [tex]S_{1} -S_{2}[/tex] is [tex]a(t-1)[/tex]
from the above the difference will increase with the time
As the water drops fall, their velocity increases due to the force of gravity, which causes the distance between each subsequent drop to increase.
Explanation:The response to the student's question deals with the notion of acceleration due to gravity. As the water drops fall, they are accelerated by gravity, which means their velocity (speed) increases over time. If we consider two subsequent droplets, the second drop begins its descent 1.0 seconds after the first. Therefore, when the second drop begins to fall, the first drop has already accelerated for 1.0 seconds. This causes the distance between the two drops to increase as they fall.
Learn more about acceleration due to gravity here:https://brainly.com/question/33911719
#SPJ3
What is a constellation as astronomers define it today? What does it mean when an astronomer says, "I saw a comet in Capricorn last night?"
A constellation, in astronomy, is a conventional grouping of stars, whose position in the night sky is apparently invariable. The peoples, generally of ancient civilizations, decided to link them through imaginary strokes, thus creating virtual silhouettes on the celestial sphere. From 1928, the International Astronomical Union (UAI) decided to officially regroup the celestial sphere into 88 constellations with precise limits, such that every point in the sky would be within the limits of a figure. When an astronomer says he saw a comet in Capricorn last night, it means that he saw a comet in the direction of the constellation of Capricorn.
A runner in a relay race runs 20 m north, turns around and runs south for 30 m, then turns north again and runs 40 m. The entire run took 30 seconds. Draw a sketch. What was the average speed of the runner?
Answer:
3.33m/s
Explanation:
The total distance that he runs is
30m north + 30m south + 40m north = 100 m
If the entire run takes 30 seconds then the average speed is distance over a unit of time
100 / 30 = 3.33 m/s
You can draw a sketch of a line going north 2 spaces, then going south by 3 spaces, finally going north again by 4 spaces
A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?
Answer:
31 m/s
Explanation:
As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion
[tex]s_m = gt_m^2/2[/tex]
[tex]25 = 9.8t_m^2/2[/tex]
[tex]t_m^2 = 50/9.8 = 5.1[/tex]
[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]
For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s
The information on a one-gallon paint can is that the coverage, when properly applied, is 440 ft2. One gallon is 231 in3. What is the average thickness of the paint in such an application
Answer:
t= 0.00364 in
Explanation:
Given that
The volume of the paint ,V= 231 in³
The surface area ,A = 440 ft²
We know that
1 ft = 12 in
That is why
A= 440 x 12 x 12 in²
A= 63360 in²
Lets take the thickness of the paint = t in
We know that
V= A t
[tex]t=\dfrac{V}{A}[/tex]
[tex]t=\dfrac{231}{63360}\ in[/tex]
t= 0.00364 in
Therefore the thickness ,t= 0.00364 in
To find the average thickness of the paint, divide the volume of paint by the area covered. This results in an average thickness of approximately 0.003645 inches, or about 0.000304 feet when the paint is applied as per the can's instructions.
Explanation:To find the average thickness of the paint applied, we need to calculate the volume of paint used per unit of area covered. The volume of one gallon of paint is 231 cubic inches, and the coverage is 440 square feet. To convert the coverage to square inches, we multiply by the number of square inches in a square foot, which is 144 (12 inches × 12 inches).
The total coverage in square inches is 440 ft2 × 144 in2/ft2 = 63,360 in2. We can then find the average thickness by dividing the volume of paint by the area covered: Thickness (in inches) = Volume (in cubic inches) / Area (in square inches).
This gives us an average thickness of 231 in3 / 63,360 in2, which simplifies to approximately 0.003645 inches. This can also be converted to feet by knowing that 1 inch is equal to 1/12 of a foot, so the average paint thickness is roughly 0.003645/12 feet when applied as instructed on the paint can.
Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes around a different level curve at twice the speed of Car 1. How much larger will the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration
Answer:
The radius of the curve that Car 2 travels on is 380 meters.
Explanation:
Speed of car 1, [tex]v_1=65\ km/h[/tex]
Radius of the circular arc, [tex]r_1=95\ m[/tex]
Car 2 has twice the speed of Car 1, [tex]v_2=130\ km/h[/tex]
We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
According to given condition,
[tex]\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}[/tex]
[tex]\dfrac{65^2}{95}=\dfrac{130^2}{r_2}[/tex]
On solving we get :
[tex]r_2=380\ m[/tex]
So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.
Final answer:
For Car 2 to have the same centripetal acceleration while traveling at twice the speed of Car 1, it must travel on a curve with a radius that is four times larger, which would be 380 meters.
Explanation:
The centripetal acceleration (ac) of a car going around a level curve at a constant speed is given by the formula ac = v2 / r, where v is the speed of the car and r is the radius of the circular path. If Car 1 is traveling at 65 km/h and has a centripetal acceleration on a curve with a radius of 95 m, and Car 2 is traveling at twice the speed of Car 1, we are asked to find the required radius of the curve for Car 2 to maintain the same centripetal acceleration.
To keep the same centripetal acceleration for Car 2, which is moving at twice the speed, the radius r2 must be increased proportionally to the square of the speed ratio. Since Car 2 is traveling at twice the speed, the radius must be increased by a factor of 22, or 4 times larger than that for Car 1. Therefore, if Car 1's radius is 95 m, Car 2's radius needs to be 95 m * 4, which is 380 m.
A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)
a. The time at which the maximum height is achieved is functionsequation editor s.
b. The maximum height achieved by the projectile is functionsequation editor m.
c. The time when the projectile hits the ground is functionsequation editor s.
d. The range of the projectile is functionsequation editor m.
e. The speed of the projectile on impact with the ground is functionsequation editor m/s.
Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
[tex]t=\frac{usin\alpha }{g}[/tex]
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
[tex]t=40sin30/9.81\\t=2.0secs[/tex]
b. the expression for the maximum height is expressed as
[tex]H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m[/tex]
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
[tex]R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m[/tex]
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°
As we learned in class if a material’s crystal structure is known a theoretical density, ????, can be computed from a tiny fundamental unit using the formula ???? = ???????? ????c???????? Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with its experimentally measured density of 7.87 g/cm^3.
Answer : Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].
Explanation :
Nearest neighbor distance, r = [tex]0.124nm=1.24\times 10^{-8}cm[/tex] [tex](1nm=10^{-7}cm)[/tex]
Atomic mass (M) = 55.85 g/mol
Avogadro's number [tex](N_{A})=6.022\times 10^{23} mol^{-1}[/tex]
For BCC = Z = 2
Given density = [tex]7.87g/cm^3[/tex]
First we have to calculate the cubing of edge length of unit cell for BCC crystal lattice.
For BCC lattice : [tex]a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.24\times 10^{-8}cm}{\sqrt{3}})^3=2.35\times 10^{-23}cm^3[/tex]
Now we have to calculate the density of unit cell for BCC crystal lattice.
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell (for BCC = 2)
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
Now put all the values in above formula (1), we get
[tex]\rho=\frac{2\times (55.85g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (2.35\times 10^{-23}Cm^3)}=7.89g/Cm^{3}[/tex]
From this information we conclude that, the given density is approximately equal to the given density.
Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].
A uniform line charge extends from x = - 2.6 cm to x = + 2.6 cm and has a linear charge density of = 5.5 nC/m.(a) Find the total charge.Find the electric field on the y axis at the following distances.(b) y = 4 cm(c) y = 12 cm
Answer with Explanation:
We are given that
x=-2.6 cm to x=2.6 cm
Linear charge density=[tex]\lambda=5.5nC/m=5.5\times 10^{-9} C/m[/tex]
[tex]1nC=10^{-9} C[/tex]
Length of wire=[tex]2.6-(-2.6)=2.6+2.6=5.2cm[/tex]
Length of wire=[tex]\frac{5.2}{100}=0.052m[/tex]
1 m=100 cm
We know that
a.Linear charge density=[tex]\frac{Q}{L}[/tex]
Where Total charge =Q
Length=L
Total charge,Q=[tex]\lambda L[/tex]
Using the formula
Total charge,Q=[tex]5.5\times 10^{-9}\times 0.052=2.86\times 10^{-10}C[/tex]
b.y=4 cm=[tex]\frac{4}{100}=0.04m[/tex]
1 m=100 cm
Electric field on the y-axis is given by
[tex]E=\frac{2\lambda}{4\pi\epsilon_0 y}(\frac{L}{\sqrt{4y^2+L^2}}[/tex]
[tex]\frac{1}{4\pi\epsilon_0}=9\times 10^9 Nm^2/C^2[/tex]
Using the formula
[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.04\times \sqrt{4(0.04)^2+(0.052)^2}}[/tex]
[tex]E=1348.8N/C[/tex]
c.y=12 cm=[tex]\frac{12}{100}=0.12m[/tex]
Using the formula
[tex]E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.12\times \sqrt{4(0.12)^2+(0.052)^2}}[/tex]
[tex]E=174.7N/C[/tex]
To find the total charge of the uniform line charge, multiply the linear charge density by the length of the line. To find the electric field at a certain distance on the y axis, use the formula for electric field due to a line charge.
Explanation:(a) To find the total charge of the line, we need to multiply the linear charge density by the length of the line:
Charge = (Linear charge density)*(Length)
Convert the length to meters:
Length = (2.6 cm + 2.6 cm) = 0.052 m
Total Charge = (5.5 nC/m)*(0.052 m) = 0.286 nC
(b) To find the electric field at y = 4 cm, we can use the formula:
Electric Field = (Linear charge density)/(2πε₀y)
Convert the linear charge density to C/m:
Charge density = 5.5 nC/m = 5.5 x 10-9 C/m
Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.04 m))
(c) To find the electric field at y = 12 cm, we can use the same formula:
Electric Field = (5.5 x 10-9 C/m)/(2πε₀(0.12 m))
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the transverse speed and acceleration of an element of the string at t = 0.160 s for the point on the string located at x = 1.40 m.Your response differs from the correct answer by more than 10%. Double check your calculations. m/sm/s2(b) What are the wavelength, period, and speed of propagation of this wave?msm/s
Explanation:
(a) It is known that equation for transverse wave is given as follows.
y = [tex](0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)[/tex]
Now, we will compare above equation with the standard form of transeverse wave equation,
y = [tex]A sin(kx + \omega t)[/tex]
where, A is the amplitude = 0.09 m
k is the wave vector = [tex]\frac{\pi}{11}[/tex]
[tex]\omega[/tex] is the angular frequency = [tex]4\pi[/tex]
x is displacement = 1.40 m
t is the time = 0.16 s
Now, we will differentiate the equation with respect to t as follows.
The speed of the wave will be:
v(t) = [tex]\frac{dy}{dt}[/tex]
v(t) = [tex]A \omega cos(kx + \omega t)[/tex]
v(t) = [tex](0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)[/tex]
v(t) = -0.84 m/s
The acceleration of the particle in the location is
a(t) = [tex]\frac{dv}{dt}[/tex]
a(t) = [tex]-A \omega 2sin(kx + \omega t)[/tex]
a(t) = [tex]-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)[/tex]
a(t) = -9.49 [tex]m/s^{2}[/tex]
Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 [tex]m/s^{2}[/tex] .
(b) Wavelength of the wave is given as follows.
[tex]\lambda = \frac{2\pi}{k}[/tex]
[tex]\lambda = (frac{2\pi}{\frac{\pi}{11}) [/tex]
[tex]\lambda[/tex] = 22 m
The period of the wave is
T = [tex]\frac{2 \pi}{\omega}[/tex]
T = [tex]\frac{2 \pi}{4 \pi}[/tex]
= 0.5 sec
Now, we will calculate the speed of propagation of wave as follows.
v = [tex]\frac{\lambda}{T}[/tex]
= [tex]\frac{22 m}{0.5 s}[/tex]
= 44 m/s
therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.
To find the transverse speed and acceleration of an element on a string given its wave function and specific values for time and position, we differentiate the wave function with respect to time. The first derivative gives us speed, and the second derivative provides acceleration. To determine the wave's wavelength, period, and propagation speed, we compare the wave function to its standard form and use the wave number and angular frequency.
Explanation:To determine the transverse speed and acceleration of a string element for the given wave function y = (0.090 m) sin (px/11 + 4pt) at t = 0.160 s and x = 1.40 m, we need to take the first and second derivatives of the wave function with respect to time. We also need to convert the given wave function into SI units for consistency.
First, let's tackle the transverse speed, which is given by the first derivative of the displacement y with respect to time t. Acceleration is given by the second derivative with respect to time.
(a) At t = 0.160 s and x = 1.40 m:
Vy = ∂y/∂t = 4p(0.090m)cos(px/11 + 4pt)
Acceleration, ay = ∂^2 y/∂t^2 = -16π^2(0.090m)sin(px/11 + 4pt)
Now plug in the values of x and t to get the numerical measure of speed and acceleration.
(b) To find the wavelength, period, and wave propagation speed, compare the wave function's format to the standard form y(x, t) = A sin (kx - ωt), where k is the wave number related to wavelength (λ) by k = 2π/λ and ω is the angular frequency related to period (T) by ω = 2π/T. The wave speed (v) is given by v = λ/T.
Identify k and ω from the wave function and calculate the wavelength, period, and wave speed.
By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zero in what configuration?
Answer: at when distance r = infinity.
Explanation: The formulae for the electric potential of an electric charge to an arbitrary point is given by the formulae below
V = q/4πεr
V = electric potential (volts)
q = magnitude of electric charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)
Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).
Same goes for electric potential energy between two charges, the formulae is given below as
W = q1 *q2/4πεr
W= electric potential energy
q1 = magnitude of first charge.
q2 = magnitude of second charge
ε = permittivity of free space
r = distance between arbitrary point and charge.
Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.
Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)
A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20 cm. (a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. (Enter the magnitude of the force.) Entry field with correct answer 1.44e-7 N This force is Entry field with correct answer attractive repulsive (b) Now the balls are moved so that as they hang, the distance between their centers is only 5 cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42
Answer:
a) F = 1.44 10⁻⁷ N, the charges are of the same sign the force is repulsive
b) F₂ / F₁ = 16
Explanation:
a) When the balls are touched the load is distributed evenly between the two balls, therefore when separating each ball has a load of
q₁ = q₂ = ½ 10¹⁰ 1.6 10⁻¹⁹ C
q₁ = q₂ = 0.8 10⁻⁹ C
As the charges are of the same sign the force is repulsive
To calculate the force let's use Coulomb's law
F = k q₁ q₂ / r²
F = 8.99 10⁹ 0.8 10⁻⁹ 0.8 10⁻⁹ / 0.20²
F = 1.44 10⁻⁷ N
b) Let us seek strength for the distress of
r₂ = 0.05 m
F₂ = 8.99 10⁹ 0.8 0.8 10⁻¹⁸ / 0.05²
F₂ = 2,301 10⁻⁶ N
b) The relationship between these two forces is
F₂ / F₁ = 23.01 10⁻⁷ / 1.44 10⁻⁷
F₂ / F₁ = 16
The falling object in Example 2 satisfies the initial value problem dv/dt =9.8−(v/5), v(0) =0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?
Answer:
a. [tex]t=19.56 s[/tex]
b.[tex]d=718.34[/tex]
Explanation:
The solution to the differential equation
[tex]\dfrac{dv}{dt}=9.8-\dfrac{v}{5}[/tex]
is the exponential function
[tex]v(t)=ce^{-0.2t}+49[/tex]
and we find [tex]c[/tex] from the initial condition [tex]v(0)=0:[/tex]
[tex]0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49[/tex]
Therefore, we have
[tex]v(t)=-49e^{-0.2t}+49[/tex]
[tex]\boxed{ v(t)=49(1-e^{-0.2t})}[/tex]
Part A:
The maximum velocity that the object can reach is 49 (which the maximum value [tex]v(t)[/tex] can have).
Now, 98% of 49 is 48.02; therefore,
[tex]48.02=49(1-e^{-0.2t})[/tex]
[tex]0.98=1-e^{-0.2t}[/tex]
[tex]e^{-0.2t}=0.02[/tex]
[tex]\boxed{t=19.56 s}[/tex]
Part B:
The distance traveled is the integral of the speed:
[tex]d=\int_0^{19.56}v(t)*dt[/tex]
[tex]d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt[/tex]
[tex]d=49[t+5e^{-0.2t}]_0^{19.56}[/tex]
[tex]\boxed{d=718.34}[/tex]
To find the time that must elapse for the object to reach 98% of its limiting velocity, we need to solve the differential equation. We can then find the distance the object falls by integrating the velocity function with respect to time.
Explanation:(a) Finding the time to reach 98% of the limiting velocityTo find the time it takes for the object to reach 98% of its limiting velocity, we need to solve the differential equation. First, we separate the variables by writing it as:
dv / (9.8 - (v/5)) = dt
Next, we integrate both sides:
∫ (1 / (9.8 - (v/5))) dv = ∫ dt
After evaluating the integrals, we can solve for v:
v = 49 - 49e^(-t/5)
Substituting v with 0.98 times the limiting velocity (which is 49), we can solve for t:
49 - 49e^(-t/5) = 0.98 * 49
Solving this equation will give us the time it takes for the object to reach 98% of its limiting velocity.
(b) Finding the distance the object fallsTo find the distance the object falls, we need to integrate the velocity function, v, with respect to time:
∫ v dt
By evaluating the integral, we can calculate the distance the object falls in the time found in part (a).
Learn more about Differential equations here:https://brainly.com/question/33814182
#SPJ3
If instead the distance between the moon and the planet were 7 times as large (no change in mass), what would the magnitude of the force be?
Answer:
Reduced by 49 times
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:
[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]
where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.
Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times
To solve the problem we should know about Newton's Law of gravity.
What is Newton's Law of gravity?
According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]
[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]
Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².
The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
Given to us,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]distance between the moon and the planet is 7 timesAssumption
Let's assume that the distance between the moon and the planet is d.
Values
As it is given that there is no change in the mass of the moon or the planet, therefore,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]Also, it is given that the distance between them changes to 7 times, therefore,
distance between the moon and the planet =7dNewton's Law of gravitySubstitute the value Newton's Law of gravity,
[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]
Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
Learn more about Newton's Law of gravity:
https://brainly.com/question/1017661