Answer:
4.8063m
Explanation:
Horzontal range is given by the formula;
R=(u²sin2θ)/g
u=7m/s, θ=53°, g=9.8m/s
[tex]R=\frac{7^{2}sin2*53 }{9.8}[/tex]
[tex]R=\frac{7^{2}sin106 }{9.8}[/tex]
[tex]R=\frac{49*sin2*53 }{9.8}[/tex]
[tex]R=\frac{47.102 }{9.8}[/tex]
R=4.8063m
A rock thrown at an angle of 53 degrees with an initial speed of 7.0 m/s from a 50-m-height cliff will hit the ground approximately 8.3 meters away from the base of the cliff.
Explanation:The problem regards the range of a projectile which is given by the formula R = (v²/g) * sin(2*Theta), where v is the initial velocity, g is the acceleration due to gravity, and Theta is the launch angle. Given the initial velocity v = 7.0 m/s, launch angle Theta = 53 degrees, and g = 9.8 m/s², it's a matter of substituting these values into the range formula: R = ((7.0 m/s)² / 9.8 m/s²) * sin(2 * 53 degrees).
After performing the calculations, you obtain that the rock will hit ground approximately 8.3 meters away from the base of the cliff.
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Kelli weighs 435 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back and releases it when the seat is 1.00 m above the ground. Assume that air resistance is negligible.A) How fast is Kelli moving when the swing passes through its lowest position? B) If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction?
Answer:A 3.57m/s
B Workdone= 246.44J
Explanation:Given: change in heigt= 1-0.36=0.64, mass=435/9.8= 44.4Kg
PE=mgh=44.4×9.8×0.64=278.4J
Let PE=KE=1/2mv2
278.4=44.4/2×v^2
278.4=22.2v^2
V^2=278.4/22.4
V^2=12.79
V=3.57m/s
B. Velocity at the bottom of the swing is 2m/s
KE=44.4×0.36×2=-31.96J
Workdone=PE+KE=278.4-31.96=246.44J
Kelli's speed when the swing passes through its lowest position is 4.14 m/s. The work done on the swing by friction is 905 N.
Explanation:To determine the speed of Kelli when the swing passes through its lowest position, we can use the principle of conservation of mechanical energy. At the highest point, Kelli has potential energy and no kinetic energy. At the lowest point, the potential energy is zero and the entire mechanical energy is in the form of kinetic energy. Therefore, the speed of Kelli at the lowest position can be found using the equation:
Kinetic energy = Mechanical energy - Potential energy
We know that the mechanical energy is the same at both points, so we can write:
0.5 mv^2 = mg
Where m is the mass of Kelli, v is her speed, g is the acceleration due to gravity, and h is the difference in height between the highest and lowest points of the swing. Rearranging the equation, we have:
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 * 9.8 m/s^2 * 1.36 m) = 4.14 m/s
Therefore, Kelli is moving at a speed of 4.14 m/s when the swing passes through its lowest position.
To find the work done on the swing by friction, we can use the work-energy principle. The work done by friction can be calculated using the equation:
Work = Change in kinetic energy
Given that the initial velocity of the swing is 2.0 m/s and the final velocity is 0 m/s due to friction, the change in kinetic energy is equal to the initial kinetic energy. Therefore, the work done on the swing by friction is:
Work = 0.5 * m * (initial velocity)^2 = 0.5 * 435 N * (2.0 m/s)^2 = 905 N
An astronaut standing on a platform on the Moon drops a hammer. If the hammer falls 6.0 meters vertically in 2.7 seconds, what is its acceleration?a. 1.6m/s²
b. 4.4m/s²
c. 2.2m/s²
d. 9.8m/s²
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Displacement, s = 6 m
Time, t = 2.7 s
Substituting
s = ut + 0.5 at²
6 = 0 x 2.7 + 0.5 x a x 2.7²
a = 1.65 m/s²
Acceleration is 1.65 m/s²
Option A is the correct answer.
A scientific _________ is a proposed explanation for a fairly narrow set of phenomena, usually based on prior experience, scientific background knowledge, preliminary observations, and logic.
Answer:
A scientific hypothesis
Explanation:
A scientific hypothesis is an idea or explanation for a fairly narrow set of phenomena that you then test through research, experimentation, experience, scientific background knowledge, preliminary observations, and logic. They are the initial building block in the scientific method and they are also beyond a wild guess but less than a well-established theory.
Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is moving upward with increasing speed, your weight will be:_______.a) greater than your normal weight at rest.b) equal to your normal weight at rest.c) less than your normal weight at rest.
Answer: The answer is A.
Explanation:
Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is moving upward with increasing speed, your weight will be greater than your normal weight at rest.
To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force F⃗ acting between two charges. The magnitude F of the force between two charges q1 and q2 depends on the product of the charges and the square of the distance r between the charges: F=k|q1q2|r2, where k=1/(4πϵ0)=8.99×109N⋅m2/C2. The direction of the force is along the line connecting the two charges. If the charges have the same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and like charges repel. Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words, if a charge q experiences a force F⃗ , the electric field E⃗ at that point is E⃗ =F⃗ q
Answer:
Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved
Explanation:
Let the charges be q1 and q2 and the distance between the charges be 'd'
Mathematical representation of coulombs law will be;
F1=kq1q2/d²...(1)
Where k is the electrostatic constant.
If q1 and q2 is doubled and the distance halved, we will have;
F2 = k(2q1)(2q2)/(d/2)²
F2 = 4kq1q2/(d²/4)
F2 = 16kq1q2/d²...(2)
Dividing equation 1 by 2
F1/F2 = kq1q2/d² ÷ 16kq1q2/d²
F1/F2 = kq1q2/d² × d²/16kq1q2
F1/F2 = 1/16
F1 = 1/16F2
This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved
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Explanation:
Coulomb's law describes the electrostatic force between charges, and the electric field is the mechanism that causes this force. The force depends on the product of the charges and the square of the distance between them. The electric field is defined as the force per unit charge and describes the influence of a charge on the surrounding space.
Coulomb's law describes the electrostatic force between two charges.
The force is given by the equation F=k|q1q2|r2, where k=8.99x10^9 N⋅m2/C2.
The direction of the force is along the line connecting the two charges, and it is repulsive for like charges and attractive for opposite charges.
The electric field is the intermediate mechanism that causes the force between charges. It is defined as the force per unit charge and can be calculated as E=F/q. The electric field describes the influence of a charge on the space around it.
For example, if there is a positive charge, it will create an electric field that points away from it. If a different positive charge is placed in this electric field, it will experience a repulsive force because the field pushes it away. On the other hand, if a negative charge is placed in the electric field, it will experience an attractive force because the field pulls it towards the positive charge.
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A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?
A.)There is no electric field on the interior of the conducting sphere.
B.)The interior field points in a direction parallel to the exterior field.
C.)The interior field points in a direction opposite to the exterior field.
D.)The interior field points in a direction perpendicular to the exterior field.
Answer:
The answer is A. There is no electric field on the interior of the conducting sphere.
Explanation:
A solid conducting sphere in a uniform electric field will exert force on the charges in the sphere to redistribute themselves in such a way that both the charges and the field inside the sphere would vanish.
| The electric field 5.0 cm from a very long charged wire is (2000 N/C, toward the wire). What is the charge (in nC) on a 1.0-cm-long segment of the wire?
Answer:
The charge is 0.056 nC.
Explanation:
Given that,
Electric field = 2000 N/C
Distance = 5.0 cm
We need to calculate the charge density
Using formula of charge density
[tex]E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}[/tex]
[tex]\lambda=2\pi\times\epsilon_{0}\times r\times E[/tex]
Put the value into the formula
[tex]\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000[/tex]
[tex]\lambda=5.56\times10^{-9}\ C/m[/tex]
We need to calculate the charge in 1.0 cm
Using formula of charge
[tex]Charge = \lambda\times\text{length of segment}[/tex]
[tex]Charge =5.56\times10^{-9}\times1.0\times10^{-2}[/tex]
[tex]Charge=0.056\times10^{-9}\ C[/tex]
[tex]Charge=0.056\ nC[/tex]
Hence, The charge is 0.056 nC.
The charge on a 1.0-cm-long segment of the wire is 2000 nanocoulombs (nC).
Explanation:The charge on a 1.0-cm-long segment of the wire can be determined using the formula:
Electric field strength = force per unit charge
Since we know the electric field strength and the length of the wire segment, we can rearrange the formula to solve for the charge:
Charge = Electric field strength x Length of wire segment
Plugging in the given values, the charge on the 1.0-cm-long segment of the wire is 2000 N/C x 1.0 cm = 2000 nC.
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A lunar vehicle is tested on Earth at a speed of 10 km/h. How fast would it have to be moving on the moon to have the same momentum it had during the test?
Answer:
10km/h
Explanation:
We are given that
Speed of lunar vehicle on Earth,v=10km/h
We have to find the speed of vehicle on moon when it had same momentum on both planet.
Let m be the mass of lunar vehicle and u be the velocity of lunar vehicle on moon
Mass does not vary with position of object on planet.
Therefore, mass of lunar vehicle remains same on earth and moon.
We know that
Momentum of object,P=mv
According to question
[tex]mv=mu[/tex]
[tex]v=u[/tex]
[tex]u=10km/h[/tex]
Hence, the lunar vehicle moving with velocity 10 km/h on moon.
Are graded receptor potentials always depolarizing? Do graded receptor potentials always make it easier to induce action potential?
Answer: Yes,graded receptor potential always depolarize.
Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.
Explanation:
A university campus has 200 classrooms and 400 faculty offices. The classrooms are equipped with 12 fluorescent tubes, each consuming 110 W, including the electricity used by the ballasts. The faculty offices, on average, have half as many tubes. The campus is open 240 days a year. The classrooms and faculty offices are not occupied for an average of 4 h a day, but the lights are kept on. If the unit cost of electricity is $0.115/kWh, determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.
Final answer:
The campus will save approximately $58,124.80 per year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.
Explanation:
To determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods, we need to calculate the annual energy consumption and cost.
The classrooms consume 12 fluorescent tubes, each consuming 110 W. Since the faculty offices have half as many tubes, they consume 6 tubes on average.
First, let's calculate the annual energy consumption for the classrooms:
Energy consumption per classroom per day = (12 tubes) × (110 W/tube) × (4 h) = 5,280 Wh
Total energy consumption for all classrooms per day = (5,280 Wh/classroom) × (200 classrooms) = 1,056,000 Wh
Annual energy consumption for classrooms = (1,056,000 Wh/day) × (240 days) = 253,440,000 Wh = 253,440 kWh
Now, let's calculate the annual energy consumption for the faculty offices:
Energy consumption per office per day = (6 tubes) × (110 W/tube) × (4 h) = 2,640 Wh
Total energy consumption for all faculty offices per day = (2,640 Wh/office) × (400 offices) = 1,056,000 Wh
Annual energy consumption for faculty offices = (1,056,000 Wh/day) × (240 days) = 253,440,000 Wh = 253,440 kWh
Adding the energy consumption of the classrooms and faculty offices, the campus consumes a total of 506,880 kWh per year. To determine the cost savings, we need to multiply this by the unit cost of electricity:
Cost savings = (506,880 kWh) × ($0.115/kWh) = $58,124.80
Therefore, the campus will save approximately $58,124.80 per year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.
What four processes make up the simple ideal Rankine cycle? Group of answer choices Constant heat addition in a boiler, Isothermal compression in a pump, constant heat rejection in a condenser and Isentropic expansion in a turbine Constant heat addition in a boiler, Isentropic compression in a pump, constant heat rejection in a condenser and Isentropic expansion in a turbine Constant heat addition in a boiler, Isentropic compression in a pump, constant heat rejection in a condenser and Isobaric expansion in a turbine Constant pressure addition in a boiler, Isentropic compression in a pump, constant heat rejection in a condenser and Isentropic expansion in a turbine
Answer:
Isentropic compression in a pump, Constant heat addition in a boiler, Isentropic expansion in a turbine, constant heat rejection in a condenser
Explanation:
Ideal rankine cycle schematic diagram and T-S diagram is shown in below figure
The process 1-2 is isentropic compression is taking place in pump.
The process 2-3 is constant heat addition, heat addition takes places at boiler.
The process 3-4 is isentropic expansion, is taking place is turbine.
The process 4-1 is constant heat rejection , takes place in condenser.
Two objects, one having three times the mass of the other, are dropped from the same height in a vacuum. At the end of their fall, their velocities are equal because the force of gravity is the same for both objects. Is this not true?
Answer:
For two or more bodies of different mass released from height in a vacuum have the same velocity but varying force
Explanation:
Consider a body H with initial velocity (u) and final velocity V undergoing acceleration a and covering a distance( s)
From Network equation of motion it can be seen that
V^2=u^2+2as
From this it can be seen that velocity is not dependent on the the masses of the body.
Rather it depends on acceleration due to gravity which is a constant for both of the body
In a vacuum, objects of different masses will have the same acceleration due to gravity and their velocities will increase at the same rate during free fall. However, at the end of their fall, their velocities will not be equal because the object with greater mass will have greater inertia.
Explanation:In a vacuum, where there is no air resistance, both objects experience the same acceleration due to gravity, regardless of their mass. This means that their velocities will increase at the same rate during free fall.
However, at the end of their fall, their velocities will not be equal because the object with greater mass will have greater inertia, and therefore require a greater force to reach the same velocity as the lighter object.
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The heat caused by the friction of moving engine parts and the explosion of gasoline in the cylinders is removed through the vehicle's cooling system.
A) TrueB) False
Answer:
True
Explanation:
Most vehicle's cooling system contain a coolant that flows through certain passages in the engine, as the coolant moves through those passages it absorb heat from moving engine parts and the explosion of gasoline in the cylinders. the coolant is then stored back in a radiator which is responsible for transferring the heat from the coolant to the environment(air).
A mysterious device found in a forgotten laboratory accumulates charge at a rate specified by the expression q(t) = 9 − 10t C from the moment it is switched on. (a) Calculate the total charge contained in the device at t = 0. (b) Calculate the total charge contained at t = 1 s. (c) Determine the current flowing into the device at t = 1 s, 3 s, and 10 s.
At t = 0 the total charge contained in the device is 9 C. At t = 1 the total charge contained in the device is -1 C. The current flowing into the device at any given instant is -10 A.
Explanation:The current, I, in a device can be determined using the rate of change of charge, q, with respect to time, t. Hence the current is equal to the derivative of the charge with respect to time. First, evaluate the expression q(t) = 9 - 10t for t = 0, 1, 3 and 10.
At t = 0, q(0) = 9 - 10*(0) = 9 C.At t = 1, q(1) = 9 - 10*(1) = -1 C.To find the current, we take the derivative of q with respect to time, dq/dt = -10. This is the current flowing into the device at any moment in time. Therefore, at t = 1 s, 3 s, and 10 s, the current is -10 A.Learn more about Electric Charge and Current here:https://brainly.com/question/3478825
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For t = 1 s, the total charge is -1 C. The current at t = 1 s, 3 s, and 10 s is constant at -10 A.
The charge q(t) contained in the mysterious device as a function of time is given by the expression q(t) = 9 - 10t C. To answer the questions, we will perform some calculations.
(a) Total charge at t = 0: By substituting t = 0 into the expression q(t) = 9 - 10t, we get q(0) = 9 C.
(b) Total charge at t = 1 s: Substituting t = 1 into the expression q(t) = 9 - 10t, we get q(1) = 9 - 10(1) = -1 C.
(c) Current at various times: The current I is the rate of change of charge with respect to time. This is found by differentiating q(t) with respect to t, which yields I(t) = -10 A (since the derivative of a constant is 0 and the derivative of -10t with respect to t is -10). The current flowing into the device at t = 1 s, 3 s, and 10 s is thus constant at -10 A.
Note that the negative sign indicates the direction of current flow (i.e., charge leaving the device).
A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The acceleration of the block is_______.
Answer:
[tex]1m/s^2[/tex]
Explanation:
Mass of block=10 kg
Applied horizontal force =F=20 N
Friction force=f=10 N
We have to find the acceleration of block.
Net force=Applied horizontal force-friction force
[tex]ma=F-f[/tex]
Where F= Horizontal force
f=Friction force
m=Mass of object
a=Acceleration of object
[tex]10a=20-10=10[/tex]
[tex]a=\frac{10}{10}=1 m/s^2[/tex]
Hence, the acceleration of the block=[tex]1m/s^2[/tex]
The acceleration of the block is 1 m/s².
What is acceleration?Acceleration can be defined as the rate of change of velocity
To calculate the acceleration of the block, we use the formula below.
Formula:
ma = F-F'............... Equation 1Where:
m = mass of the blocka = acceleration of the blockF = horizontal force applied to the blockF' = friction force.Make a the subject of the equation
a = (F-F')/m............. Equation 2From the question,
Given:
F = 20 NF' = 10 Nm = 10 kgSubstitute these values into equation 2
a = (20-10)/10a = 10/10a = 1 m/s²Hence, The acceleration of the block is 1 m/s².
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An astronaut notices that a pendulum which took 2.45 s for a complete cycle of swing when the rocket was waiting on the launch pad takes 1.25 s for the same cycle of swing during liftoff.
What is the acceleration (m/s²) of the rocket?(Hint: Inside the rocket, it appears that g has increased.)
Answer:
2.84 g's with the remaining 1 g coming from gravity (3.84 g's)
Explanation:
period of oscillation while waiting (T1) = 2.45 s
period of oscillation at liftoff (T2) = 1.25 s
period of a pendulum (T) =2π. [tex]\sqrt{\frac{L}{a} }[/tex]
where
L = lengtha = accelerationtherefore the ration of the periods while on ground and at take off will be
[tex]\frac{T1}{T2}[/tex] =(2π [tex]\sqrt{\frac{L}{a1} }[/tex] ) / (2π[tex]\sqrt{\frac{L}{a2} }[/tex])
where
a1 = acceleration on ground while waitinga2 = acceleration during liftoff[tex]\frac{T1}{T2}[/tex] = [tex]\frac{\sqrt{\frac{L}{a1} }}{\sqrt{\frac{L}{a2} }}[/tex]
squaring both sides we have
[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{\frac{L}{a1} }{\frac{L}{a2} }[/tex]
[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{a2}{a1}[/tex]
assuming that the acceleration on ground a1 = 9.8 m/s^{2}
[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{a2}{9.8}[/tex]
a2 = 9.8 x [tex](\frac{T1}{T2})^{2}[/tex]
substituting the values of T1 and T2 into the above we have
a2 = 9.8 x [tex](\frac{2.45}{1.25})^{2}[/tex]
a2 = 9.8 x 3.84
take note that 1 g = 9.8 m/s^{2} therefore the above becomes
a2 = 3.84 g's
Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.
To find the rocket's acceleration during liftoff, we can use the change in the period of a pendulum. By equating the expressions for the pendulum length before and during liftoff, we determine that the acceleration is approximately 37.62 m/s².
To find the rocket's acceleration during liftoff using the pendulum's period, we can use the formula for the period of a pendulum:
T = 2π √(L/g)
When the rocket was on the launch pad, the period T₀ was 2.45 s and the acceleration due to gravity g₀ was 9.80 m/s². Let's denote the pendulum length as L. We can write:
2.45 = 2π √(L/9.80)
During liftoff, the period T changed to 1.25 s. The new effective acceleration due to gravity inside the accelerating rocket is g. We can write:
1.25 = 2π √(L/g)
Since the length L of the pendulum did not change, we can equate the expressions for L and solve for g. Rearrange both equations to solve for L:
L = (T₀ / 2π)² × 9.80L = (1.25 / 2π)² × gEquate the two expressions for L:
(2.45 / 2π)² × 9.80 = (1.25 / 2π)² × g
Solve for g:
g = ( (2.45)² / (1.25)² ) × 9.80
= (6.0025 / 1.5625) × 9.80
≈ 37.62 m/s²
Therefore, the rocket's acceleration during liftoff is 37.62 m/s².
At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magnitude of the gravitational field strength at this location? (1) 0.20 kg/N (2) 9.8 m/s2 (3) 5.0 N/kg (4) 25 000 Nkg
Answer:
3. 5.0N/kg
Explanation:
Gravitational field strength = gravitational force/mass of astronaut = 350N/70kg = 5.0N/kg
Final answer:
The magnitude of the gravitational field strength where a 70.0-kilogram astronaut experiences a gravitational force of 350 newtons is 5.0 N/kg, calculated by dividing the force by the mass.
Explanation:
To calculate the magnitude of the gravitational field strength at a location where a gravitational force of 350 newtons acts on a 70.0-kilogram astronaut, we divide the force by the mass of the astronaut. This gives us the gravitational field strength (g) which can be represented by the equation g = F/m, where F is the gravitational force and m is the mass of the object experiencing the force.
In this case, the calculation will be g = 350 N / 70.0 kg, which equals 5.0 N/kg. Therefore, the magnitude of the gravitational field strength at this location is 5.0 N/kg, which corresponds to choice (3).
In the morning, when the temperature is 286 K , a bicyclist finds that the gauge pressure in his tires is 404 kPa . That afternoon she finds that the pressure in the tires has increased to 422 kPa .
Part A What is the afternoon temperature?
Final answer:
Using the gas law principle and given pressure values, the afternoon temperature of the bicyclist's tires, after an increase in pressure to 422 kPa from 404 kPa, is calculated to be approximately 299 K.
Explanation:
To find the afternoon temperature after the tire pressure increases, we use the gas law principle that describes the relationship between pressure and temperature when the volume and amount of the gas remain constant. Assuming ideal behavior, the relevant equation is the proportionality between pressure and temperature given as P1/T1 = P2/T2.
In this case, the initial pressure (P1) is 404 kPa, the initial temperature (T1) is 286 K, and the final pressure (P2) is 422 kPa. To find the final temperature (T2), we rearrange the formula to T2 = (P2 x T1) / P1.
Plugging in the values:
T2 = (422 kPa x 286 K) / 404 kPa
T2 = 1207962 K kPa / 404 kPa
T2 ≈ 299 K
Therefore, the afternoon temperature when the pressure in the tires has increased to 422 kPa is approximately 299 K.
A metal object is suspended from a spring scale. The scale reads 943 N when the object is suspended in air, and 799 N when the object is completely submerged in water.
a. Find the volume of the object. The acceleration of gravity is 9.8 m/s². Answer in units of m³.
b. Find the density of the metal. Answer in units of kg/m³.
To find the volume of the metal object, subtract the weight of the object submerged in water from its weight in air. Then, use the buoyant force formula to calculate the volume. The density of the metal cannot be determined without knowing the mass of the object.
Explanation:In order to find the volume of the metal object, we need to first determine the buoyant force that acts on it when it is submerged in water. The difference between the scale readings in air and water represents the buoyant force exerted by the water on the object. The weight of the object in air is 943 N and in water, it is 799 N. The buoyant force is the difference between these two values, which is 143 N.
The buoyant force is equal to the weight of the water displaced by the object. Therefore, the volume of the object can be calculated using the buoyant force and the density of water. The density of water is approximately 1000 kg/m³. We can use the formula:
Volume = Buoyant Force / (Density of Water * Acceleration due to Gravity)
Substituting the given values, we have:
Volume = 143 N / (1000 kg/m³ * 9.8 m/s²)
Simplifying the equation, we find that the volume of the object is 0.0146 m³.
Now, to find the density of the metal, we can divide the mass of the object by its volume. Since the mass of the object is not given in the question, we cannot provide a specific answer to part (b) without this information.
An iron cannon ball and a bowling ball are dropped at the same time from the top of a building. At the instant before the balls hit the sidewalk, the heavier cannonball has a greater?A. velocity. B. aceleration. C. kinetic energy. D. all of these are the same for the two balls.
Answer:
C. kinetic energy
Explanation:
Free Fall
When objects are dropped in free air (no friction), they are attracted to the Earth's surface by the force of gravity. Objects start from zero speed and gradually increase it because of the effect of the acceleration of gravity whose value can be considered as constant [tex]g=9.8\ m/s^2.[/tex]
The final speed of the object at time t is
[tex]v_f=g.t[/tex]
Note that the final speed does not depend on the mass of the object. The kinetic energy is
[tex]\displaystyle k=\frac{mv_f^2}{2}[/tex]
The kinetic energy depends on the speed and the mass, so heavier objects dropped from the same height will have more kinetic energy. Let's analyze the options given in the question .
A. As shown above, the speed (or magnitude of the velocity) is the same regardless of the object's mass, so the heavier cannonball and the iron cannon will reach the ground at the same speed. Incorrect option
B. For every falling object in free air, the only acceleration is the acceleration of gravity. This option is not correct either
C. The kinetic energy is greater for the heavier cannonball because it has more mass, as discussed above. Correct option
D. Only the C. option is correct. This is not
Final answer:
At the instant before the balls hit the ground, the cannonball with greater mass will have greater (C) kinetic energy, while acceleration and velocity will be the same for both balls due to the constant acceleration of gravity. Hence, (C) is the correct option.
Explanation:
The question revolves around the physics concept of objects falling under gravity. When two objects, such as an iron cannon ball and a bowling ball, are dropped from the same height without any air resistance, the acceleration due to gravity acts equally on both.
Regardless of their masses, their acceleration remains constant at ~9.8 m/s². Therefore, the cannonball does not have a greater velocity or acceleration compared to the bowling ball. What differentiates them at the instant before impact is their kinetic energy, which is dependent on mass.
The heavier cannonball will indeed have greater kinetic energy because kinetic energy is calculated using the equation KE = 1/2 m v², where 'm' represents mass and 'v' represents velocity. Since the masses are different and velocities are the same, the cannonball with the greater mass will have greater kinetic energy.
The process of cutting lines of magnetic flux with a conductor, and generating voltage, is the basisi of alternator and generator operation.True / False.
Answer:
True
Explanation:
Alternators and generators are devices which convert mechanical energy into electrical energy, they consist of a magnet and a conductor wound about an iron called a core. When the conductor cuts across the lines of flux of the magnet an electromotive force is induced thereby generating current, either the conductor or the magnet has to be moving for this to happen and when the direction of motion changes the direction of electromotive force also changes.
Final answer:
True. The process of cutting lines of magnetic flux with a conductor and generating voltage is the basis of alternator and generator operation.
Explanation:
True.
The process of cutting lines of magnetic flux with a conductor and generating voltage is the basis of alternator and generator operation. This is known as electromagnetic induction, which is the process of inducing an electromotive force (emf) or voltage with a change in magnetic flux. When a coil is rotated in a magnetic field, it produces an alternating current emf, which is the basic construction of a generator.
A car travels in a straight line covering a total distance of 90.0 miles in 60.0 minutes. Which one of the following statements concerning this situation is necessarily true?
Answer:
E) The average velocity of the car is 90.0 miles per hour in the direction of motion.
Explanation:
From the available options below:
A) The velocity of the car is constant.
B) The acceleration of the car must be non-zero.
C) The first 45 miles must have been covered in 30.0 minutes.
D) The speed of the car must be 90.0 miles per hour throughout the entire trip.
E) The average velocity of the car is 90.0 miles per hour in the direction of motion.
The average velocity of an object is the total distance covered by the object while moving in a particular direction per total time taken to cover the distance.
From the information given, during the course of the journey, the velocity may vary and the acceleration may be zero at some point if the driver of the car decides to stop over.
However, the average velocity for the journey must be 90.0 miles per hour in the direction of the motion in order for the car to be able to cover 90 miles in 60 minutes.
The correct option is E.
The primary advantage of a bipolar scheme is that when all the voltages are added together after a long transmission, there should be a total voltage of ____.
Answer:
There should be a total voltage of zero (0)
Explanation:
In bipolar encoding (multilevel binary), there are three voltage levels, positive, negative, and zero. The voltage level for one data element is at zero, while the voltage level for the other element alternates between positive and negative.
However, the primary advantage of a bipolar scheme is that when all the voltages are added together after a long transmission, there should be a total voltage of zero.
A bicycle travels 141 m along a circular track of radius 30 m. What is the angular displacement in radians of the bicycle from its starting position?
Answer: 4.7rad
Explanation:
Angular displacement =s/r
Where s=distance traveled
r=radius
Angular displacement =141m/30m
Angular displacement =4.7rad.
The angular displacement in radians of the bicycle from its starting position is 4.7 radians.
From the given information;
the distance at which the bicycle travels = 141 mthe radius of the circular path = 30 mThe angular displacement θ in a circular path can be expressed by using the equation:
[tex]\mathbf{\theta = \dfrac{s}{r}}[/tex]
[tex]\mathbf{\theta = \dfrac{141 }{30}}[/tex]
[tex]\mathbf{\theta =4.7 \ radians}[/tex]
Learn more about angular displacement here:
https://brainly.com/question/13902407?referrer=searchResults
A simple harmonic oscillator consists of a spring of constant k = 100 N/m and a block of mass 2.0 kg. When t = 1.0 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s. (a) Findthe amplitude of the oscillations? (b) Determinethe location (x) of the block at t = 0? (c) What is the velocity v(t) of the block at t = 0?
Answer:
a) 0,18 m b) 0,18 m c) 0 m/s
Explanation:
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward.
Part A
What equation describes this motion as a function of time?
Part B
At what time will the spring stretch to its maximum length at first time?
Part C
At what time will the spring shrink to its minimum length at first time?
Answer:
[tex]y(t)=0.28sin(36.02599t)[/tex]
0.0436 s
0.1308 s
Explanation:
A = Amplitude = 28 cm
m = Mass = 0.235 kg
k = Spring constant = 305 N/m
The equation which describes motion as a function of time is given by
[tex]y(t)=Asin(\omega t)[/tex]
Angular speed is given by
[tex]\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \omega=\sqrt{\dfrac{305}{0.235}}\\\Rightarrow \omega=36.02599\ rad/s[/tex]
The equation is
[tex]\mathbf{y(t)=0.28sin(36.02599t)}[/tex]
Maximum length will be at amplitude
Amplitude is given by
[tex]A=Asin(\omega t)[/tex]
here
[tex]\omega t=\dfrac{\pi}{2}\\\Rightarrow t=\dfrac{\pi}{2\times 36.02599}\\\Rightarrow t=0.0436\ s[/tex]
The time to stretch to maximum length is 0.0436 s
At minimum length
[tex]y(t)=-A\\\Rightarrow -A=Asin\omega t[/tex]
[tex]\omega t=\dfrac{3\pi}{2}\\\Rightarrow t=\dfrac{3\pi}{2\omega}\\\Rightarrow t=\dfrac{3\pi}{2\times 36.02599}\\\Rightarrow t=0.1308\ s[/tex]
The time will the spring shrink to its minimum length at first time is 0.1308 s
The answer explains the motion equation, time for maximum and minimum lengths, and provides a detailed solution.
Part A: The equation that describes this motion as a function of time for a vertical spring with a mass hanging from it is y(t) = A * sin(2πft + φ), where A is the amplitude, f is the frequency, t is time, and φ is the phase angle.
Part B: The time when the spring stretches to its maximum length for the first time is half a period after passing the equilibrium point, so Tmax = 0.5 / f.
Part C: The time when the spring shrinks to its minimum length for the first time is one period after passing the equilibrium point, so Tmin = 1 / f.
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10−21−21 NN?
Answer:
There are 756.25 electrons present on each sphere.
Explanation:
Given that,
The force of repression between electrons, [tex]F=3.33\times 10^{-21}\ N[/tex]
Let the distance between charges, d = 0.2 m
The electric force of repulsion between the electrons is given by :
[tex]F=k\dfrac{q^2}{r^2}[/tex]
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}[/tex]
[tex]q=1.21\times 10^{-16}\ C[/tex]
Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :
q = ne
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}[/tex]
n = 756.25 electrons
So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Earth's axis of a person standing on the surface of the Earth at a latitude of 71.9^{\circ}71.9 ∘ ?
The centripetal acceleration of someone at a latitude of 71.9 degrees on Earth's surface is calculated based on the velocity due to Earth's rotation, adjusted by the cosine of the latitude angle, demonstrating how this acceleration compares to Earth's gravitational pull.
Explanation:The question asks about the centripetal acceleration of a person standing on Earth's surface at a specific latitude and how this compares to Earth's gravitational acceleration, g. To find the centripetal acceleration (ac) at a latitude of 71.9 degrees, we use the fact that at the equator, where the velocity due to Earth's rotation is maximal, ac is purely in the direction of g and reduces with latitude. The reduction factor is the cosine of the latitude angle because the actual velocity component contributing to centripetal acceleration is the component of the Earth's rotational velocity at the surface that is perpendicular to the axis of rotation, which is maximized at the equator and zero at the poles.
To calculate it: The earth rotates once every 24 hours, giving us a rotational velocity at the equator. We can derive velocity (v) at the equator using the formula v = 2πr / T, where r is Earth's radius (approximately 6380 km) and T is the period of rotation (86400 seconds). The equatorial velocity can then be used to calculate centripetal acceleration at the equator with ac = v² / r, which is found to be a fraction of g, as shown in discussions. At a latitude of 71.9 degrees, this acceleration is reduced by the factor cos(71.9°), illustrating how ac changes with latitude.
Thus, to express the centripetal acceleration at this latitude as a multiple of g, one would calculate the acceleration as derived for the equator and adjust it by the factor cos(71.9°). This allows us to understand the reduced effect of Earth's rotation on the apparent weight of an object as the latitude increases, which is crucial in understanding phenomena such as the variation in gravity and the operation of pendulum clocks at different latitudes.
Will Mark As Brainliest Plz Help Unit 1 Physics B Test
1. Two charges of magnitude +q are separated by a distance d. Assuming there are no changes to anything else, but the charges are separated by 3 times their current distance, what will happen to the Coloumb force between them?
The new Coloumb force will be 1/9th the original
The new Coloumb force will be 1/3rd of the original
The new Coloumb force will be 3 times the original
The new Coloumb force will be 9 times the original
2. In the nucleus of an atom, two protons are separated by a distance of 10^-13 m. What is the Coloumb force between them? The charge of a proton is + 1.602 x 10^-19 C.
0.023 N
0.078 N
0.004 N
0.051 N
3. How many electrons must be added to an object to give it an overall net charge of -3.0 x 10^-7 C? The charge of an electron is -1.602 x 10^-19 C. 1.9 x 10^12 electrons
3.1 x 10^13 electrons
2.0 x 10^10 electrons
1.4 x 10^14 electrons
4. What is the net Coloumb force on the +1e charge? A charge of e is 1.602 x 10^-19 C.
3.91 x 10^-11 C to the right
1.62 x 10^-10 C to the right
2.02 x 10^-9 C to the left
4.97 x 10^-10 C to the left
5. A point p is located 56 cm to the right away from a point charge of 6.7 x 10^-10 C. What is the direction and magnitude of the electric field at point p?
19.21 N/C
23.7 N/C
11.5 N/C
13.8 N/C
6. A point q is located 151 cm directly above a charge of -5.5 x 10^-12 C. What is the direction and magnitude of the electric field at point q?
1.35 N/C, pointing down
3.84 N/C, pointing up
2.99 N/C, pointing up
2.17 N/C, pointing down
7. What is the net electric force at point x in the picture below?
0.29 N/C to the left
0.09 N/C to the left
0.37 N/C to the right
0.14 N/C to the right
8. A point charge of charge Q is located some distance d away from a location p. If nothing changes, but p is moved twice as far away, what will happen to the electric field at point p?
The electric field will be 1/2 of what it was
The electric field will be 1/3 of what it was
The electric field will be 1/4 of what it was
The electric field will be 1/5 of what it was
9. If an atom has 13 protons and is currently electrically neutral, what must happen to give the same atom a positive charge of +2e?
2 electrons must be removed
3 electrons must be added
1 proton must be removed
2 protons must be removed
What is polarization?
10. When the positive and negative charges in an object are separated
The creation of magnetic poles
The charging of an object by placing a charged object against a neutral one
The creation of charge by removing charges through friction
Answer: Did you get the answers? Im struggling homie.
Explanation:
Answer:
1. 1/9th the original
2. 0.023 N
3. 1.9 x 10^12 electrons
4. 1.62 x 10^-10 C to the right
5. 19.21 N/C
6. 2.17 N/C
7. Not 0.37 N/C
8. 1/4 of what it was
9. 2 electrons must be removed
10. Polarization: when the positive and negative charges in an object are separated
Explanation: Just took the test
Two cellists, one seated directly behind the other in an orchestra, play the same 220-Hz note for the conductor who is directly in front of them. What is the smallest non-zero separation that produces constructive interference? Take the speed of sound to be 343 m/s.
Answer:
Constructive interference defination:
''Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes''
Constructive interference occurs at integer multiples of the wavelength of the wave. The lowest incidence occurs at the wavelength.
As we know,
wavelength * frequency = velocity
wavelength = v/f
= (343 m/s) / (220 1/s)
= 1.56 m