Answer:
it will move by d = 8.00 m in x direction before it will turn back
Explanation:
Here the initial velocity of particle is along +x direction
it is given as
[tex]v_i = 4.90 m/s[/tex]
now its acceleration is given as
[tex]a_x = -1.50 m/s^2[/tex]
[tex]a_y = 3.00 m/s^2[/tex]
now when it turns back then the velocity in x direction will become zero
so we will say
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 4.90^2 = 2(-1.50) d[/tex]
[tex] d = 8.00 m[/tex]
so it will move by d = 8.00 m in x direction before it will turn back
Final answer:
The particle moves 8.01 m in the x direction before it turns around, determined by using kinematic equations with the given initial velocity and constant acceleration.
Explanation:
To determine how far the particle moves in the x direction before turning around, we need to find the point at which its velocity in the x direction is zero. Since the particle starts with an initial velocity of 4.90 m/s in the x direction and has a constant acceleration of ax = -1.50 m/s², we can use the kinematic equation v = u + at to find the time when the velocity becomes zero:
v = 0 m/s (the velocity at the turnaround point)
u = 4.90 m/s (initial velocity)
a = -1.50 m/s² (constant acceleration)
Solving for the time (t) gives:
0 = 4.90 + (-1.50)t
t = 4.90 / 1.50
t = 3.27 s
Now using the equation s = ut + (1/2)at² to find the displacement in the x direction:
s = 4.90 m/s ⋅ t + (1/2)(-1.50 m/s²) ⋅ t²
s = 4.90 ⋅ 3.27 + (0.5) ⋅ (-1.50) ⋅ (3.27)²
s = 8.01 m
The particle moves 8.01 m in the x direction before turning around.
A potato is shot out of cylinder at an angle of 17 degrees above the horizontal with an initial speed of 20 m/s. What is its maximum height?
Answer:
Maximum height, h = 1.74 meters
Explanation:
It is given that,
A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.
Initial speed with which the potato is shot out, u = 20 m/s
We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :
[tex]h=\dfrac{u^2sin^2\theta}{2g}[/tex]
Where
[tex]\theta[/tex] = angle between the projectile and the surface
g = acceleration due to gravity
[tex]h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}[/tex]
h = 1.74 m
or h = 1.74 meters
Hence, this is the required solution.
Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.48~\mathrm{Watts/m^2}1.48 Watts/m 2 . If a radar unit leaks 10.0~\text{Watts}10.0 Watts of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Recall that Watts = Joules/second = power = energy per unit time. Assume that the power of the electromagnetic waves spreads uniformly in all directions (i.e. spreads out over the area of a sphere) and use the formula for the surface area of a sphere.
Answer:
0.733 m
Explanation:
The maximum safe intensity for human exposure is
[tex]I= 1.48 W/m^2[/tex]
Intensity is defined as the ratio between the power P and the surface irradiated A:
[tex]I=\frac{P}{A}[/tex]
For a source emitting uniformly in all directions, the area is the surface of a sphere of radius r:
[tex]A=4 \pi r^2[/tex]
So
[tex]I=\frac{P}{4\pi r^2}[/tex]
In this case, we have a radar unit with a power of
P = 10.0 W
So we can solve the previous equation to find r, which is the distance at which a person could be considered to be safe:
[tex]r=\sqrt{\frac{P}{4\pi I}}=\sqrt{\frac{10.0 W}{4 \pi (1.48 W/m^2)}}=0.733 m[/tex]
A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express your answer in degrees Celsius to one decimal place.
Answer:
Final temperature of the aluminum = 41.8 °C
Explanation:
We have the equation for energy
E = mcΔT
Here m = 55 g = 0.055 kg
ΔT = T - 27.5
Specific heat capacity of aluminum = 921.096 J/kg.K
E = 725 J
Substituting
E = mcΔT
725 = 0.055 x 921.096 x (T - 27.5)
T - 27.5 = 14.31
T = 41.81 ° C = 41.8 °C
Final temperature of the aluminum = 41.8 °C
An aluminum block of 55.0 g at an initial temperature of 27.5 °C absorbs 725 J of heat. Using the formula for heat transfer, we calculate that the final temperature of the aluminum block is approximately 36.8 °C.
Explanation:This question can be answered using the formula for heat transfer Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat of aluminum and ΔT is the change in temperature. We've been given the mass, heat amount, and initial temperature. We also know that the specific heat of aluminum is 0.897 J/g°C (derived from the Table 5.1 and Table 9.1).
So the equation becomes 725 = 55 * 0.897 * (T_final -27.5). Solving for T_final, we get that the final temperature of the aluminum block is approximately 36.8 °C. Please note that this is a simplification as it doesn't take into account any heat losses to the surrounding environment.
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Formula One race cars are capable of remarkable accelerations when speeding up, slowing down, and turning corners. At one track, cars round a corner that is a segment of a circle of radius 95 m at a speed of 68 m/s. What is the approximate magnitude of the centripetal acceleration, in units of g?
Answer:
Centripetal acceleration of the car is (4.96 g) m/s²
Explanation:
It is given that,
Radius of circle, r = 95 m
Speed of the car, v = 68 m/s
We need to find the centripetal acceleration. It is given by :
[tex]a_c=\dfrac{v^2}{r}[/tex]
So, [tex]a_c=\dfrac{(68\ m/s)^2}{95\ m}[/tex]
[tex]a_c=48.67\ m/s^2[/tex]
Since, g = 9.8 m/s²
So,
[tex]a_c=(4.96\ g)\ m/s^2[/tex]
So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.
A 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s. What is the induced emf in the wire? O 3.18 V O 6.36 v O 4.78 V O 12.3V
Answer:
Induced emf of the wire is 6.36 Volts.
Explanation:
It is given that,
Length of the wire, l = 75 cm = 0.75 m
Magnetic field, B = 0.53 T
Velocity, v = 16 m/s
The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :
[tex]\epsilon=Blv[/tex]
[tex]\epsilon=0.53\ T\times 0.75\ m\times 16\ m/s[/tex]
[tex]\epsilon=6.36\ V[/tex]
So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".
Answer:
6.36
Explanation:
If a 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s, the induced emf in the wire is 6.36.
Each croquet ball in a set has a mass of 0.52 kg. The green ball, traveling at 11.3 m/s, strikes the blue ball, which is at rest. Assuming that the balls slide on a frictionless surface and all collisions are head-on, find the final speed of the blue ball in each of the following situations: a) The green ball stops moving after it strikes the blue ball. Answer in units of m/s.
Answer:
11.3 m/s
Explanation:
Momentum is conserved:
Total momentum before collision = total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.52 kg) (11.3 m/s) + (0.52 kg) (0 m/s) = (0.52 kg) (0 m/s) + (0.52 kg) v
v = 11.3 m/s
Suppose a disk with constant angular velocity has rotational kinetic energy 1280 J. If the moment of inertia of the disk is 35 kg-m^2, then what is its angular velocity? (a) 7.604 rad/s (b) 8.552 rad/s (c) 10.12 rad/s (d) 6.818 rad/s (e) 9.952 rad/s (f) 8.935 rad/s f
Answer:
Angular velocity of the disk is 8.552 rad/s
Explanation:
It is given that,
Rotational kinetic energy, KE = 1280 J
The moment of inertia of the disk, I = 35 kg m²
We have to find the angular velocity of the disk. In rotational mechanics the kinetic energy of the disk is given by :
[tex]KE=\dfrac{1}{2}I\omega^2[/tex]
[tex]\omega=\sqrt{\dfrac{2KE}{I}}[/tex]
[tex]\omega=\sqrt{\dfrac{2\times 1280\ J}{35\ kgm^2}}[/tex]
[tex]\omega=8.552\ rad/s[/tex]
Hence, the angular velocity of the disk is 8.552 rad/s.
An electric car accelerates for 8.0 s by drawing energy from its 320-V battery pack. During this time, 1300 C worth of electrons pass through the battery pack. (a) How many electrons are moved through the battery during this 8.0 s acceleration time? (b) How much energy transfer does this constitute? (c) Find the minimum horsepower rating of the car.(746 W = 1 hp)
(a) [tex]8.13\cdot 10^{-21}[/tex]
The magnitude of the charge of one electron is
[tex]q=1.6\cdot 10^{-19}C[/tex]
Here the total amount of charge that passed through the battery pack is
Q = 1300 C
So this total charge is given by
Q = Nq
where
N is the number of electrons that has moved through the battery
Solving for N,
[tex]N=\frac{Q}{q}=\frac{1300 C}{1.6\cdot 10^{-19} C}=8.13\cdot 10^{-21}[/tex]
(b) [tex]4.16\cdot 10^5 J[/tex]
First, we can find the current through the battery, which is given by the ratio between the total charge (Q = 1300 C) and the time interval (t = 8.0 s):
[tex]I=\frac{Q}{t}=\frac{1300 C}{8.0 s}=162.5 A[/tex]
Now we can find the power, which is given by:
[tex]P=VI[/tex]
where
V = 320 V is the voltage
I = 162.5 A is the current
Subsituting,
[tex]P=(320 V)(162.5 A)=52,000 W[/tex]
And now we can find the total energy transferred, which is the product between the power and the time:
[tex]E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J[/tex]
(c) 69.7 hp
Now we have to convert the power from Watt to horsepower.
We know that
1 hp = 746 W
So we can set up the following proportion:
1 hp : 746 W = x : 52,000 W
And by solving for x, we find the power in horsepower:
[tex]x=\frac{1 hp \cdot 52,000 W}{746 W}=69.7 hp[/tex]
Final answer:
The electric car moves 8.12 times [tex]10^{21}[/tex]electrons through the battery during 8 seconds of acceleration, which constitutes a 416,000 J energy transfer. The minimum horsepower rating of the car is approximately 69.7 hp.
Explanation:
To generate an accurate answer for the question posed by the student, we must apply the principles of physics regarding electric current and energy.
Part (a)
The number of electrons moved through the battery is calculated using the charge of an electron (1.60 times [tex]10^{-19}[/tex]Coulombs). For 1300 C of charge:
Number of electrons = Total charge / Charge of one electron = 1300 C / (1.60 times [tex]10^{-19}[/tex] C)
Number of electrons = 8.12 times [tex]10^{21}[/tex] electrons
Part (b)
The energy transfer is found by multiplying the total charge by the voltage of the battery:
Energy = Charge times Voltage = 1300 C times 320 V
Energy = 416,000 J (Joules)
Part (c)
The minimum horsepower rating of the car can be found by converting the energy transfer to watts and then to horsepower:
Power (in watts) = Energy / Time = 416,000 J / 8.0 s
Power = 52,000 W
Horsepower = Power (in watts) / 746 W/hp = 52,000 W / 746 W/hp
Horsepower
= 69.7 hp (rounded to one decimal place)
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80 m/s. What is the momentum of the 202 kg car afterwards? (Unit-kg*m/s)
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s[/tex]
p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
The potential difference between two parallel plates is 227 V. If the plates are 6.8 mm apart, what is the electric field between them? O S.0x10 N/C O 28 x 10* N/C O 4.1 x 10 N/C O 3.3 x 10 N/C
Answer:
[tex]E=3.3\times 10^4N/C[/tex]
Option D is the correct answer.
Explanation:
Electric field, E is the ratio of potential difference and distance between them.
Potential difference, V = 227 V
Distance between plates = 6.8 mm = 0.0068 m
Substituting,
[tex]E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C[/tex]
Option D is the correct answer.
A satellite with mass 500 kg is placed in a circular orbit about Earth (Mass= 5.98 x 10^24 kg), radius = (6.4 x 10^6), a distance of 1500 km above the surface. (a) what is the force gravity acting on satellite? (b) what is the satellite’s acceleration? (c) what is the satellite’s orbital speed?
Explanation:
a) F = GmM / r²
F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²
F = 3200 N
b) F = ma
3200 = 500a
a = 6.4 m/s²
c) a = v² / r
640 = v² / (6.4×10⁶ + 1.5×10⁶)
v = 7100 m/s
A person walks across a 60 m wide room with a constant speed of 2 m/s, and they run back with a constant speed of 6 m/s. What is their average speed for the trip? a. 2 m/s b 3m/s c. 4 m/s d. 5 m/s 2 e 6 m/s
Answer:
Option B is the correct answer.
Explanation:
Total distance traveled = 60 + 60 = 120 m
Time taken to walk across [tex]=\frac{60}{2}=30s[/tex]
Time taken to run back [tex]=\frac{60}{6}=10s[/tex]
Total time taken = 30 + 10 = 40 s
Average speed = Total distance traveled / Total time taken
Average speed [tex]=\frac{120}{40}=3m/s[/tex]
Option B is the correct answer.
15. Give an example for some particles or waves that are moving faster than light in everyday life 16. What is a laser? 17. What is an oscilloscope? 18. What does it means practically that nothing is faster than light in vacuum? 19. What is vacuum?
Answer:
15.Radiowave
16.laser is device that generates an intense beam of other electromagnetic radiation by emission of photons from excited atoms.
17.this is a laboratory instrument commonly used to display and analyse the waveformof electronic signals.
19. this is a space entirely devoid of matter.
Julie is walking around a track at a 2m/s for some exercise. She then decides to start jogging so she accelerates at a rate of 0.5m/s² for 3 seconds. How far did Julie travel from the time she started to accelerate to the end of the 3 seconds?
Answer:
The distance is 8.25 m.
Explanation:
Given that,
Speed = 2 m/s
Acceleration = 0.5 m/s^2
Time = 3 sec
We need to calculate the distance
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
Where, u = initial velocity
a = acceleration
t = time
Put the value in the equation
[tex]s=2\times3+\dfrac{1}{2}\times0.5\times3^2[/tex]
[tex]s=8.25\ m[/tex]
Hence, The distance is 8.25 m.
Julie's distance at the end of the 3 seconds is 8.25 m.
To calculate the distance traveled by Julie at the end ]of 3 seconds, we use the formula below.
Formula:
s = ut+at²/2.............. Equation 1Where:
s = distance traveled by Julie after 3 secondsu = Initial velocityt = timea = acceleration.From the question,
Given:
u = 2 m/sa = 0.5 m/s²t = 3 secondsSubstitute these given values into equation 1
s = 2(3)+(0.5×3²)/2s = 6+2.25s = 8.25 mHence, Julie's distance at the end of the 3 seconds is 8.25 m.
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An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s^2. Determine the orbital period of the satellite.
Answer:
Orbital period, T = 2.02 hours
Explanation:
It is given that, an artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s². We have to find the orbital period (T) of the satellite.
Firstly, calculating the distance between Earth and satellite. The acceleration due to gravity is given by :
[tex]a=\dfrac{GM}{r^2}[/tex]
G = universal gravitational constant
M = mass of earth
[tex]r=\sqrt{\dfrac{GM}{a}}[/tex]
[tex]r=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{6.03\ m/s^2}}[/tex]
r = 8126273.3 m..........(1)
Now, according to Kepler's third law :
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
Putting the value of r from equation (1) in above equation as :
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.97\times 10^{24}}\times (8126273.3)^3[/tex]
[tex]T^2=53202721.01\ s[/tex]
T = 7294.01 seconds
Since, 1 hour = 3600 seconds
Converting seconds to hour we get :
So, T = 2.02 hour
So, the orbital period of the satellite is 2.02 hours.
Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods close to each, but without contact, and briefly ground the the neutral rod by touching it with your hand. show answer Correct Answer What would be resulting charge (if any) on the initially neutral rod
Answer:
I think it will be half of the initial charge
Explanation:
Because we know, the resulting charge will be q1+q2/2, since one is neutral so the charge will be half q/2
What is a simple pendulum?
Answer:
A pendulum is weight suspended from a pivot so that it can swing freely
During takeoff, an airplane climbs with a speed of 150 m/s at an angle of 45 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. The sun is shining directly overhead. How fast is the shadow of the plane moving along the ground
Answer:
106.1 m/s
Explanation:
The shadow of the plane is moving at the same velocity as the horizontal component of the airplane's velocity.
The horizontal component of the airplane's velocity is
[tex]v_x = v cos \theta[/tex]
where
v = 150 m/s is the velocity of the airplane
[tex]\theta=45^{\circ}[/tex] is the angle between the airplane's velocity and the horizontal
Substituting,
[tex]v_x = (150 m/s) cos 45^{\circ} = 106.1 m/s[/tex]
So, the shadow is moving at 106.1 m/s as well.
please explain vector addition, triangle method and parallelogram method
Start with 2 arbitrary vectors, [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex]. (pic 1)
Vectors are determined by their lengths and direction. This means that translating the vector (i.e. sliding it left/right and up/down in the plane) doesn't fundamentally change that vector. To this end, we could just as easily represent [tex]\vec v_2[/tex] as if it had originated from the tip of [tex]\vec v_1[/tex]. This "new" [tex]\vec v_2[/tex] and the "old" [tex]\vec v_2[/tex] are the same vector. (pic 2)
If we connect the origin of [tex]\vec v_1[/tex] with the tip of "new" [tex]\vec v_2[/tex], we get a new vector, and this we define as the vector sum [tex]\vec v_1+\vec v_2[/tex]. (pic 3)
We can do this other way, by first traslating [tex]\vec v_1[/tex] to the tip of [tex]\vec v_2[/tex], then connecting the origin of [tex]\vec v_2[/tex] with the tip of "new" [tex]\vec v_1[/tex]. This demonstrates that vector addition is commutative (order of the vectors being added doesn't matter - you always end up at the same terminus). The "parallelogram method" refers to how a parallelogram is traced out. (pic 4)
Multiplying a vector by -1 reverses its direction. (pic 5)
Adding [tex]\vec v_1[/tex] and [tex]-\vec v_2[/tex] works the same way as standard vector addition, giving us the new vector [tex]\vec v_1-\vec v_2[/tex]. (pic 6)
We can do the same in the reverse order, but now we get a different vector, [tex]\vec v_2-\vec v_1[/tex]. (pic 7)
These vectors have the same length but point in opposite directions. (pic 8)
But notice that we can translate the vectors [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] so that we get a vector that either starts at the tip of [tex]\vec v_2[/tex] and ends at the tip of [tex]\vec v_1[/tex] (pic 9), or starts at the tip of [tex]\vec v_1[/tex] and ends at the tip of [tex]\vec v_2[/tex] (pic 10). The "triangle method" refers to the triangles that are traced out by either vector sum [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] together with [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex].
A spring has a natural length of 8 m. If a 12-N force is required to keep it stretched to a length of 10 m, how much work W is required to stretch it from 8 m to 16 m? (Round your answer to two decimal places.)
Answer:
The work is required to stretch it from 8 m to 16 m is 192 N-m
Explanation:
Given that,
Natural length = 8 m
Force F = 12 N
After stretched,
length = 10 m
We need to calculate the elongation
[tex]x = 10-8=2\ m[/tex]
Using hook's law
The restoring force is directly proportional to the displacement.
[tex]F\propto (-x)[/tex]
[tex]F = -kx[/tex]
Where, k = spring constant
Negative sign shows the displacement in opposite direction
Now, The value of k is
[tex]k = \dfrac{F}{x}[/tex]
[tex]k = \dfrac{12}{2}[/tex]
[tex]k = 6[/tex]
When stretch the string from 8 m to 16 m.
Then the elongation is
[tex]x=16-8=8\ m[/tex]
Now, The work is required to stretch it from 8 m to 16 m
[tex]W = \dfrac{1}{2}kx^2[/tex]
Where, k = spring constant
x = elongation
[tex]W=\dfrac{1}{2}\times6\times8\times8[/tex]
[tex]W=192\ N-m[/tex]
Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m
What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40×104 m/s2? What is the magnitude of the force exerted on the ship by the artillery shell?
Answer:
Force exerted, F = 2.64 × 10⁷ Newton
Explanation:
It is given that,
Mass of the artillery shell, m = 1100 kg
It is accelerated at, [tex]a=2.4\times 10^4\ m/s^2[/tex]
We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :
F = ma
[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]
F = 26400000 Newton
or
F = 2.64 × 10⁷ Newton
So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.
Answer: The force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex] and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]
Explanation:
Force is defined as the push or pull on an object with some mass that causes change in its velocity.
It is also defined as the mass multiplied by the acceleration of the object.
Mathematically,
[tex]F=ma[/tex]
where,
F = force exerted on the artillery shell
m = mass of the artillery shell = 1100 kg
a = acceleration of the artillery shell = [tex]2.40\times 10^4m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=1100kg\times 2.40\times 10^4m/s^2\\\\F=2.64\times 10^6N[/tex]
Now, according to Newton's third law, every action has an equal and opposite reaction.
So, the force exerted on the artillery shell will be equal to the force exerted on the ship by artillery shell acting in opposite direction.
Hence, the force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex] and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]
A ball is launched at an angle of 39.8 degrees up from the horizontal, with a muzzle velocity of 6.6 meters per second, from a launch point which is 1 meters above the floor. How high will the ball be above the floor (in meters), when it is a horizontal distance of 2.7 meters away? Use 9.82 meters per second for "g".
Answer:
1.85 m
Explanation:
The horizontal velocity of the ball is
[tex]v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s[/tex]
The horizontal distance travelled is
d = 2.7 m
And since the motion along the horizontal direction is a uniform motion, the time taken is
[tex]t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s[/tex]
The vertical position of the ball is given by
[tex]y= h + u_y t - \frac{1}{2}gt^2[/tex]
where
h = 1 m is the initial heigth
[tex]u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s[/tex] is the initial vertical velocity
g = 9.82 m/s^2 is the acceleration due to gravity
Substituting t = 0.53 s, we find the height of the ball at this time:
[tex]y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m[/tex]
Answer:
To convert kilometers per hour to meters per second, perform dimensional analysis. Remember that:
1 km = 1000 m
1 hr = 3600 seconds
Using these conversion, perform dimensional analysis:
16.2 km/ hr (1000m/1km) (1hr/60 sec) = 4.5 m/s
The analysis basically just uses the conversion factors and canceling of units. The final answer is 4.5 m/s.
_________________________________________________________
Correction: That should be *(1 hr/3600 sec). The answer is still 4.5 m/s.
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Hope this helps, i did the test and this answer was right, oh and brainliest, Good luck.
An intercontinental ballistic missile goes from rest to a speed of 6.50 km/s in 60.0 s. In multiples of g, what is its acceleration?
Answer:
The acceleration is [tex]11.1g\ m/s^2[/tex]
Explanation:
Given that,
Speed [tex]v= 6.50\ km/s=6.5\times10^{-3}\ m/s[/tex]
Time t = 60.0 sec
We need to calculate the acceleration
Using formula off acceleration
[tex]a = \dfrac{\Delta v}{t}[/tex]
[tex]a=\dfrac{v_{f}-v_{i}}{t}[/tex]
We know that,
Missile goes from rest
So, Initial velocity =0
Put the value into the formula
[tex]a =\dfrac{6.50\times10^{3}}{60.0}[/tex]
[tex]a=108.33\ m/s^2[/tex]
On right hand side multiplying and dividing by g = 9.8m/s²
[tex]a=108.33\times\dfrac{g}{g}[/tex]
Put the value of g
[tex]a = \dfrac{108.33}{9.8}g\ m/s^2[/tex]
[tex]a = 11.1g\ m/s^2[/tex]
Hence, The acceleration is [tex]11.1g\ m/s^2[/tex]
A locomotive enters a station with an initial velocity of 19 m/s and slows down at a rate of .8m/s^2 as it goes through. If the station is 175 m long, how fast is it going when the nose leaves the station?
Answer:
Final velocity, v = 25.3 m/s
Explanation:
Initial velocity of a locomotive, u = 19 m/s
Acceleration of the locomotive, a = 0.8 m/s²
Length of station, d = 175 m
We need to find its final velocity (v) when the nose leaves the station. It can be calculated using the third law of motion :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad+u^2[/tex]
[tex]v^2=2\times 0.8\ m/s^2\times 175\ m+(19\ m/s)^2[/tex]
[tex]v^2=(641)\ m^2[/tex]
v = 25.31 m/s
v = 25.3 m/s
When the nose leaves the station, it will move with a velocity of 25.3 m/s. Hence, this is the required solution.
A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s. Both blocks stick together and move to the right. What is their speed after collision?
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}[/tex]
v = 2 m/s
Hence, their speed after collision is 2 m/s.
On January 22, 1943, in Spearfish, South Dakota, the temperature rose from −4.00°F to 45.0°F over the course of two minutes (the current world record for the fastest recorded temperature change). By how much did the temperature change on the Kelvin scale? HINT
Answer:
27.22 K
Explanation:
T₁ = initial temperature in fahrenheit = - 4.00 ⁰F
T₂ = final temperature in fahrenheit = 45.0 ⁰F
To convert the temperature from fahrenheit to kelvin, we can use the relation
[tex]K = \frac{F - 32}{1.8} + 273.15[/tex]
where F = Temperature in fahrenheit and K = temperature in kelvin
T'₁ = initial temperature in kelvin = (- 4.00 - 32)/1.8 + 273.15 = 253.15 K
T'₂ = final temperature in kelvin = (45.0 - 32)/1.8 + 273.15 = 280.37 K
ΔT = Change in temperature
Change in temperature on kelvin scale is given as
ΔT = T'₂ - T'₁
ΔT = 280.37 - 253.15
ΔT = 27.22 K
To find the temperature change on the Kelvin scale, convert the given temperatures from Fahrenheit to Kelvin and subtract them.
Explanation:The temperature change on the Kelvin scale can be determined by converting the given temperatures from Fahrenheit to Kelvin and then finding the difference between them.
First, convert -4.00°F to Kelvin:
273.15 K + (-4.00°F + 459.67 °F) × (5/9)
Next, convert 45.0°F to Kelvin:
273.15 K + (45.0°F + 459.67 °F) × (5/9)
Finally, subtract the two Kelvin temperatures to find the temperature change.
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An electromagnetic wave of frequency 1.90 x 10^14 Hz propagates in carbon tetrachloride with a speed of 205 x 10^8m/s. What is the wavelength of the wave in carbon tetrachloride?
The wavelength of an electromagnetic wave in carbon tetrachloride with a frequency of 1.90 x 10^14 Hz and a speed of 205 x 10^8 m/s is approximately 107.89 x 10^-6 meters or 107.89 μm.
The wavelength (λ) of an electromagnetic wave can be calculated using the formula: λ = c/f
where: -
λ is the wavelength,
c is the speed of light in the medium, and
f is the frequency of the wave.
In this case, the frequency (f) is given as 1.90 x 1014 Hz, and the speed of light in carbon tetrachloride (c) is given as 205 x 108 m/s.
λ = 205 x 108 m/s / 1.90 x 1014 Hz
λ ≈ 107.89 x 10-6 m
So, the wavelength of the electromagnetic wave in carbon tetrachloride is approximately 107.89 x 10-6 meters or 107.89 μm.
A spherical surface completely surrounds a collection of charges. Find the electric flux (with its sign) through the surface if the collection consists of (a) a single +6.60 × 10-6 C charge, (b) a single -1.30 × 10-6 C charge, and (c) both of the charges in (a) and (b).
Answer:
(a) 6.8 x 10^5 Nm^2/C
(b) 1.47 x 10^5 Nm^2/C
(c) 5.3 x 10^5 Nm^2/C
Explanation:
According to the Gauss's theorem
Electric flux = Charge enclosed / ∈0
(a) Charge enclosed = 6 x 10^-6 C
So, Electric flux = (6 x 10^-6) / (8.854 x 10^-12) = 6.8 x 10^5 Nm^2/C
(b) Charge enclosed = -1.3 x 10^-6 C
So, Electric flux = (1.3 x 10^-6) / (8.854 x 10^-12) = 1.47 x 10^5 Nm^2/C
(c) Charge enclosed = 6 x 10^-6 + (-1.3 x 10^-6) = 4.7 x 10^-6 C
So, Electric flux = (4.7 x 10^-6) / (8.854 x 10^-12) = 5.3 x 10^5 Nm^2/C
The electric flux through a spherical surface due to enclosed charges can be computed using Gauss's Law. The flux for a +6.60 x 10^-6 C charge is outward-directed, for a -1.30 x 10^-6 C charge it is inward-directed, and with both charges, the net flux is the sum of the individual fluxes.
Explanation:The student is asking about the concept of electric flux through a spherical surface that surrounds a collection of charges, which falls under the subject of Physics (specifically electromagnetism), and it is a high school- or introductory college-level question. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the enclosed electric charge. This can be calculated using the formula Φ = q/ε0, where Φ is the electric flux, q is the electric charge, and ε0 is the permittivity of free space (approximately 8.85 x 10^-12 C2/N⋅m2).
For part (a), a spherical surface surrounding a single +6.60 × 10-6 C charge would result in an outward-directed electric flux Φ = +6.60 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.
For part (b), a spherical surface surrounding a single -1.30 × 10-6 C charge would have an inward-directed electric flux Φ = -1.30 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.
For part (c), when both charges are enclosed, their net flux through the surface is the sum of the individual fluxes. Therefore the net electric flux is Φ = (+6.60 × 10^-6 C - 1.30 × 10^-6 C) / 8.85 × 10^-12 C2/N⋅m2, which simplifies to the sum of the charges divided by the permittivity of free space.
An 0.80-kilogram object tied to the end of a 2.0-meter string swings as a pendulum. At the lowest point of its swing, the object has a kinetic energy of 10.0 joules. What is the speed of the object at the instant that the string makes an angle of 50.0° with the vertical?
Answer:
3.3 m/s
Explanation:
As the object rises above the lowest point, some of the kinetic energy is converted to potential energy. From the diagram, we can see that at angle θ, the object rises to height h:
h = L - L cos θ
Conservation of energy:
KE₀ = KE + PE
KE₀ = 1/2 mv² + mgh
Substituting:
KE₀ = 1/2 mv² + mg(L - L cos θ)
Given KE₀ = 10.0 J, m = 0.80 kg, g = 9.8 m/s², L = 2.0 m, and θ = 50.0°:
10.0 = 1/2 (0.80) v² + (0.80) (9.8) (2.0 - 2.0 cos 50.0)
v = 3.32 m/s
Rounding to 2 sig-figs, the speed of the object is 3.3 m/s.
For some applications, it is important that the value of a resistance not change with temperature. For example, suppose you made a 3.80?k? resistor from a carbon resistor and a Nichrome wire-wound resistor connected together so the total resistance is the sum of their separate resistances.What value should each of these resistors have (at 0 ?C) so that the combination is temperature independent?(two answers)
To achieve a temperature independent resistance of 3.80 kΩ, you need to use a carbon resistor and Nichrome wire-wound resistor that counterbalance each other. This is possible because carbon and Nichrome have opposite temperature coefficients of resistance.
Explanation:In order for the resistance to remain constant with temperature, the carbon resistor and the Nichrome wire-wound resistor must counterbalance each other. Meaning, when one's resistance increases with temperature, the other's resistance decreases, keeping the total resistance the same. Given that carbon and Nichrome have opposite temperature coefficients of resistance, they can accomplish this task.
Generally, the resistance R of a resistor is given by the formula R = R0(1 + α(T-T0)), where α is the temperature coefficient, T is the temperature and R0 is the resistance at reference temperature T0. As the temperature increases, a positive α will increase the resistance while a negative α will decrease it.
To make the combined resistance temperature independent, the sum of the change in resistance of the carbon resistor and the Nichrome resistor should be zero. Therefore, you would set up the equation where the increase of the carbon resistance equals the decrease of the Nichrome resistance. Solving this equation will give you the exact values required for the resistances of carbon and Nichrome at 0ºC in order to have a total resistance of 3.80 kΩ.
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