Answer:
Ax =7 cm
Explanation:
Analysis:
Let A be the vector of rectangular components Ax and Ay:
Where:
Ay=7 cm : The The y-component of vector A-component of vector A
α =45 : Angle of A with the positive x-axis
Ax : The x-component of vector A
Because Ax and Ay are the components of a right triangle we apply the following formula:
[tex]tan\alpha =\frac{A_{y} }{A_{x} }[/tex]
[tex]tan45=\frac{7}{A_{x} }[/tex]
[tex]A_{x} =\frac{7}{tan45}[/tex]
Ax =7 cm
Building helps children develope their _____ skills
A. number sense and operations
B. operations and geometry
C. measurement and counting
D. measurement and geometry
Answer:
D. measurement and geometry
Explanation:
Building helps children develop their _ measurement and geometry__ skills. The measurement and geometry skill of the children by observing buildings.
Geometry is a method or tool for understanding the relations among shapes and spatial properties. Children can develop spatial reasoning and can visualize shapes in different positions (orientation) when observing a building.
Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.13; that between the heavier block and the plane is 0.31. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string
Answer:
a. a = [tex]6.41 m/s^2[/tex]
b. T = -0.81 N
Explanation:
Given,
weight of the lighter block = [tex]w_1\ =\ 3.0\ N[/tex]weight of the heavier block = [tex]w_2\ =\ 7.0\ N[/tex]inclination angle = [tex]\theta\ =\ 30^o[/tex]coefficient of kinetic friction between the lighter block and the surface = [tex]\mu_1\ =\ 0.13[/tex]coefficient of kinetic friction between the heavier block and the surface = [tex]\mu_2\ =\ 0.31[/tex]friction force on the lighter block = [tex]f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta[/tex]friction force on the heavier block = [tex]f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta[/tex]Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.
From the f.b.d. of the lighter block,
[tex]w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]
From the f.b.d. of the heavier block,
[tex]w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)[/tex]
From eqn (1) and (2), we get,
[tex]w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\[/tex]
[tex]\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\[/tex]
[tex]\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.[/tex]
part (b)
From the eqn (2), we get,[tex]T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N[/tex]
Answer:
(a) 2.793 m/s^2
(b) 0.335 N
Explanation:
Let a be the acceleration in the system and T be the tension in the string.
W1 = 3 N
W2 = 7 N
m1 = 3 /1 0 = 0.3 kg
m2 = 7 / 10 = 0.7 kg
θ = 30°
μ1 = 0.13, μ2 = 0.31
Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.
By the laws of friction
f1 = μ1 x N1
Where, N1 be the normal reaction acting on block 1.
So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N
By the laws of friction
f2 = μ2 x N2
Where, N2 be the normal reaction acting on block 2.
So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N
Apply Newton's second law for both the blocks
W1 Sin30 - T - f1 = m1 a
3 Sin 30 - T - 0.337 = 0.3 x a
1.163 - T = 0.3 a ..... (1)
W2 Sin30 + T - f2 = m2 a
7 Sin30 + T - 1.88 = 0.7 x a
1.62 + T = 0.7 a ..... (2)
By solving equation (1) and (2) we get
a = 2.793 m/s^2
(b) Put the value of a in equation (2), we get
1.62 + T = 0.7 x 2.793
T = 0.335 N
A 11400 kg lunar landing craft is about to touchdown on the
surface of the moon. acceleration due to gravity onmoon is 1.0
m/s2. At an altitude of 165 m the craft's downwardvelocity is 18.0
ms . To slow down the craft, a retrorocket isfired to provide an
upward thrust. Assuming the descent isvertical, find the magniturde
of the trhust needded to reduce thevelocity to zero at the instant
when the craft touches the lunarsurface.
Answer:
F = 22572N
Explanation:
We will first calculate the acceleration needed to achieve landing speed equals zero:
[tex]V_{f}^{2}=V_{o}^{2}+2*a*\Delta Y[/tex]
[tex]0=18^{2}+2*a*(-165)[/tex] Solving for a:
[tex]a=0.98 m/s^{2}[/tex]
Now that we have the required acceleration, we need to check all forces acting on the craft:
[tex]F_{thrust} - W = m*a[/tex] where W = 11400N; m = 11400Kg; a = 0.98m/s2
Solving the equation:
[tex]F_{thrust}=22572N[/tex]
The magniture of thrust required to reduce the lunar craft's velocity to zero can be calculated using the principles of kinematics and forces. The change in kinetic energy is calculated and equated with the work done by the retro-rocket's thrust, in order to find the required thrust.
Explanation:The physics involved in this problem is related to kinematics and forces. To calculate the amount of thrust needed, we first need to find the change in kinetic energy. The initial kinetic energy is 0.5 * 11400 kg * (18 m/s)^2 and the final kinetic energy is zero since the craft is at rest. Therefore, the change in kinetic energy is -0.5 * 11400 kg * (18 m/s)^2.
By the work-energy theorem, the work done by the retro-rocket's thrust is equal to the change in kinetic energy. Since work done is also equal to force (thrust) multiplied by distance, we can equate these two expressions to find the thrust: Thrust * 165 m = -0.5 * 11400 kg * (18 m/s)^2. Solve for thrust to get the magnitude of thrust needed.
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A race car has a speed of 80 m/s when the driver releases a drag parachute. If the parachute causes a deceleration of 4m/s^2, how far will the car travel before it stops?
Answer:
The car travels 800m
Explanation:
We will need two formulas to solve this question:
[tex]V_{f}=V_{0}+at\\X=V_{0}t+\frac{at^{2}}{2}[/tex]
We can obtain the the time by using the final velocity formula and replacing with known values:
[tex]V_{f}=V_{0}+at\\\\0\frac{m}{s}=80\frac{m}{s}-4\frac{m}{s^{2}}t\\\\4\frac{m}{s^{2}}t=80\frac{m}{s}\\t=20 s[/tex]
Since we know the time, the question will be answered by inputing all the variables needed in the distance formula:
[tex]X=V_{0}t+\frac{at^{2}}{2}\\\\X=80\frac{m}{s}(20s)-\frac{4(20s)^{2}}{2}=800m[/tex]
Final answer:
The race car will travel 800 meters before coming to a stop after deploying a drag parachute causing a deceleration of 4 m/s^2 from an initial speed of 80 m/s.
Explanation:
To calculate how far the race car will travel before it stops after deploying a drag parachute, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. Considering that the deceleration is constant, the kinematic equation we can use is v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since the car comes to a stop, v = 0 m/s. The initial speed is u = 80 m/s and the deceleration provided by the drag parachute is a = -4 m/s2 (negative because it's deceleration). Plugging these values into the equation gives us:
0 = (80 m/s)2 + 2(-4 m/s2)s
Solving for s, we find:
s = (80 m/s)2 / (2 * 4 m/s2)
s = 6400 m2/s2 / 8 m/s2
s = 800 meters
Therefore, the race car will travel 800 meters before coming to a stop after the drag parachute is deployed.
A satellite of mass 5000 kg orbits the Earth (mass = 6.0 x 1024 kg) and has a period of 6000 s. In the above problem the altitude of the satellite above the Earth's surface is
Answer:
altitude of satellite from earth surface is 7140 km
Explanation:
given data:
mass of satellite = 5000kg
altitude of satellite from earth surface can be calculated by using given relation
[tex]p^2 = \frac{(4* \pi^2)(a^3)}{(G*M)}[/tex]
where,
p - satellite's period in seconds,
a - satellite's altitude in meters,
G - gravitational constant (6.67 x 10^-11 m³/kgs²) and
M - mass of Earth 5.98 x 10^24 kg
solving for altitude we have
[tex]a^3 =\fracp{(G*M*p^2)}{(4*\pi^2)}[/tex]
[tex]a^3 = 3.637 * 10^{20}[/tex]
a = 7,138,000 meters = 7,140 km
altitude of satellite from earth surface is 7140 km
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
[tex]a=\dfrac{v-u}{t}[/tex].............(1)
Now, the second equation of motion is :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value of a in above equation as :
[tex]s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2[/tex]
[tex]s=\dfrac{t(u+v)}{2}[/tex]
[tex]u=\dfrac{2s}{t}-v[/tex]
[tex]u=\dfrac{2\times 47}{8.6}-2.3[/tex]
u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
What uniform magnetic field, applied perpendicular to a beam
ofelectrons moving at 1.30 x 106 m/s, is required to
makethe elctrons travel in a ciruclar arc of radius 0.350 m?
Answer:
The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]
Explanation:
Given that,
Speed [tex]v=1.30\times10^{6}\ m/s[/tex]
Radius = 0.350 m
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B =\dfrac{mv}{qr}[/tex]
Where, m = mass of electron
v = speed of electron
q = charge of electron
r = radius
Put the value into the formula
[tex]B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}[/tex]
[tex]B=2.11\times10^{-5}\ T[/tex]
Hence, The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]
A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car? O a. 0.72 kN O b.0.68 kN O c. 0.64 kN O d.0.76 kN O e. 0.52 kN
Answer:
c)[tex]F_{net} = 0.640 kN[/tex]
Explanation:
As we know that resultant force is the net force that is acting on the system
As per Newton's II law we know that net force is product of mass and acceleration
so we will have
[tex]F_{net} = ma[/tex]
here we know
m = 80 kg
for circular motion acceleration is given as
[tex]a_c = \frac{v^2}{R}[/tex]
[tex]a_c = \frac{40^2}{200} = 8 m/s^2[/tex]
now we have
[tex]F_{net} = 80 \times 8[/tex]
[tex]F_{net} = 640 N[/tex]
[tex]F_{net} = 0.640 kN[/tex]
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard by the observer if the source emits a 2550-Hz sound?
Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
[tex]f = f_0(\dfrac{v-v_0}{v-v_s})[/tex]
[tex]f = 2550\times (\dfrac{343+17}{343-0})[/tex]
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
A 280 g object is attached to a spring and executes simple harmonic motion with a period of 0.270 s. If the total energy of the system is 4.75 J. (a) Find the maximum speed of the object _______ m/s (b) Find the force constant of the spring_________ N/m (c) Find the amplitude of the motion_________m
Answer:
Explanation:
given,
mass of the object = 280 g = 0.28 kg
time period = 0.270 s
total energy of the system = 4.75 J
[tex]\dfrac{1}{2}\ m\ V^2 = 4.75 J[/tex]
maximum speed of the object V = [tex]\sqrt{ \dfrac{2 \times 4.75}{0.28} }[/tex]
V= 5.82 m / s
(b) force constant of the spring K = m ω²
where ω = angular frequency = 2π / T
T= time period = 0.25 s
ω = 25.13 rad / s
K = 0.28 × 25.13²
K = 176.824 N / m
(c). Amplitude of motion A = [tex]\dfrac{V}{\omega}[/tex]
= [tex]\dfrac{5.82}{25.13}[/tex]
A = 0.232 m
Aaron is flying a helicopter with an air-relative velocity of 12.5 m/s to the West. A constant wind is also blowing the helicopter towards the South at 4.55 m/s relative to the ground. What is the velocity vector of the helicopter relative to the ground? What is the speed and direction relative to the East? How long does it take to travel 1 km?
Answer:
Explanation:
Given
Velocity of helicopter relative to air 12.5 m/s to west
In vector form
[tex]V_{ha}=-12.5\hat{i}[/tex]
where [tex]V_h[/tex]= velocity of helicopter relative to ground
Also the velocity of air
[tex]V_a=4.55\hat{j}[/tex]
[tex]V_{ha}=V_h-V_a[/tex]
[tex]V_h=V_{ha}+V_a[/tex]
[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]
Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]
=13.302
For Direction
[tex]tan\theta =\frac{4.55}{-12.5}[/tex]
[tex]180-\theta =20[/tex]
[tex]\theta =160^{\circ}[/tex] relative to east
For 1 km travel it takes
[tex]t=\frac{1000}{13.302}=75.17 s[/tex]
When a particular wave is vibrating with a frequency of 4.00 Hz, a transverse wave of wavelength 60 cm is produced. Determine the speed of waves along the wire.
Answer:
Speed of the wave, v = 2.4 m/s
Explanation:
Given that,
Frequency of vibrations, f = 4 Hz
Wavelength of the transverse wave, [tex]\lambda=60\ cm=0.6\ m[/tex]
We need to find the speed of the waves along the wire. Let it is equal to v. So, the relationship is given by :
[tex]v=f\times \lambda[/tex]
[tex]v=4\times 0.6[/tex]
v = 2.4 m/s
So, the speed of waves along the wire is 2.4 m/s. Hence, this is the required solution.
Final answer:
The speed of the transverse wave on the wire, given a frequency of 4.00 Hz and a wavelength of 60 cm, is calculated to be 2.40 meters per second (m/s) using the formula Speed = Frequency × Wavelength.
Explanation:
To determine the speed of a wave, you can use the formula:
Speed = Frequency × Wavelength
In the given scenario, the frequency is 4.00 Hz, and the wavelength is 60 cm, which is equivalent to 0.60 meters (since 1 meter = 100 cm). Thus, the speed of the wave on the wire can be calculated as follows:
Speed = 4.00 Hz × 0.60 m
Speed = 2.40 m/s
Therefore, the speed of the transverse wave along the wire is 2.40 meters per second (m/s).
Which properties should an object, system, or process have in order for it to be a useful standard of measurement for a physical quantity such as length or time? (Choose all that apply) -precise-restricted availability-defined in SI units-small-reproducible
Answer:
Precise and Reproducible
Explanation:
For a system or a process to be useful a measurement standard for any physical quantity, it must be:
Precise, Precision is important in any measurement.The values of the measurement obtained should be reproducibleThe measurement may be large or smallIt may or may not be in SI unitsIt should be available easily.Answer:
precise
reproducible
Explanation:
Two balls are kicked with the same initial speeds. Ball A was kicked at the angle 20° above horizontal and ball B was kicked at the angle 75° above horizontal. What ball will have bigger speed at the highest point of their trajectory? (section 4.3) Ball A Ball B Impossible to answer without knowing their actual initial speeds. They will have the same speeds
Answer:
Ball A
Solution:
According to the question:
Two balls A and B are kicked at angle [tex]20^{\circ}[/tex] and [tex]75^{\circ}[/tex] respectively.
The component of the velocity that provides the ball with the forward movement is the horizontal component, i.e., cosine of the angle.
Since, cos[tex]20^{\circ}[/tex] is greater than cos[tex]75^{\circ}[/tex] and thus Ball A will have more speed than ball B at the top most point of the trajectory.
Final answer:
Ball A will have a bigger speed at the highest point of its trajectory in comparison to ball B, as it was kicked at a lower angle of 20°, resulting in a larger horizontal component of the initial velocity when compared to ball B kicked at 75°.
Explanation:
The question is asking about the speeds of two balls at the highest point of their trajectories when they are kicked with the same initial velocity but at different angles. To determine the speed at the highest point, we can use the principles of projectile motion. The speed of a projectile at its highest point consists only of the horizontal component of the initial velocity because the vertical component equals zero at that instant.
Since both balls A and B were kicked with the same initial speed, the horizontal component of that speed can be found using the cosine of their respective launch angles. For ball A, this component is cos(20°) multiplied by the initial speed, and for ball B, it is cos(75°) multiplied by the initial speed. Because the cosine of 20° is greater than the cosine of 75°, ball A will have a larger horizontal component of velocity and thus a higher speed at the peak of its trajectory.
A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. Part A: What is the speed of the rock just before it hits the street?
Part B: How much time elapses from when the rock is thrown until it hits the street?
Answer:
A. 33.77 m/s
B. 6.20 s
Explanation:
Frame of reference:
Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m
Initial velocity upwards v= 27 m/s
Part A. Using kinematics expression for velocities and distance:
[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]
Part B. Using Kinematics expression for distance, time and initial velocity
[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]
Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.
The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.
Explanation:Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:
Initial potential energy = final potential energy + final kinetic energy
mgh = 1/2m*v^2
where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.
Since the mass of the rock cancels out, we have:
gh = 1/2v^2
Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:
v = sqrt(2gh)
v = sqrt(2*9.8*21.0)
v = sqrt(411.6)
v = 20.3 m/s
Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.
Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:
t = 2v/g
where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.
Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:
t = 2*27.0/9.8
t = 5.5 s
Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.
A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What is the damping constant? (b) How does the amplitude vary with time? (c) Determine the critical damping constant?
For a mass-spring system with specified parameters, the damping constant is found to be 4.999 Ns/m. The amplitude reduces exponentially over time. The critical damping constant, defining the transition from underdamped to overdamped behavior, is 5 Ns/m.
Explanation:This problem pertains to a system of mass springs undergoing damped oscillations. The damping constant (b) can be calculated using the formula b = 2*m*ωd. Here ωd signifies the damped frequency and is given by ωd = ω * (1 - γ/2), where ω is the natural frequency and γ is the damping ratio of 0.2% = 0.002. The natural frequency ω = sqrt(k/m) = sqrt(12.5/0.5) = 5 rad/s. Substituting these values in the equation gives b = 2*0.5*4.999 = 4.999 Ns/m.
The amplitude (A) of a damped oscillation decreases exponentially with time due to energy loss, described by A(t) = A0 * e^(-γωt/2), wherein A0 is the initial amplitude.
The critical damping constant (bc), describing the boundary between underdamped and overdamped systems, is equal to 2*sqrt(m*k) = 2*sqrt(0.5*12.5) = 5 Ns/m.
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A particle (q = 3.0 mC, m = 20 g) has a speed of 20 m/s when it enters a region where the electric field has a constant magnitude of 80 N/C and a direction which is the same as the velocity of the particle. What is the speed of the particle 3.0 s after it enters this region?
Answer:
56 m/s
Explanation:
The electirc force applied on the particle by the field will be
F = q * E
F = 3*10^-3 * 80 = 0.24 N
This force will cause an acceleration:
F = m * a
a = F/m
a = 0.24 / 0.02 = 12 m/s^2
The equation for speed under constant acceleration is:
V(t) = V0 + a*t
V(3) = 20 + 12 * t = 56 m/s
The final speed will be 56 m/s.
The molecules in metal are usually more tightly packed than the molecules in water. Most metals will sink in water. Pick your conclusion from this list. Water is more dense than metal
Metal is more dense than water
Water and Metal have the same density
Answer:
Metal is more dense than water.
Explanation:
As we know, the molecules of the metal are tightly closed as compared to that of water and the density of a material is defined as the mass of the material per unit volume.
In a certain volume of metal, there are more numbers of molecules than that of water in the same amount of volume, therefore the density of metal is greater than that of water.
Also, according to Archimedes' principle, if there is an object in a fluid, then the buoyant force on that object is equal to the weight of the fluid that it displaces.
When the density of the object is larger than that of the fluid then it overcomes that buoyant force and sinks.
Thus, an object sinks in a fluid if its density is larger than that of the fluid and floats otherwise.
Since, the metal sink in water, it means Metal is more dense than water.
A student is driving to school on a road with a speed limit of 40mph (17.9 m/s). A stoplight ahead of her turns yellow, so she begins to slow down to come to a stop. On a dry day, she is able to make her car slow down with an acceleration of magnitude of 8.8 m/s^2. However, if the road is wet, she can only manage an acceleration with a magnitude 3.9 m/s^2. Part A:
On a dry day, how far (in meters) from the intersection would the student need to start braking in order to stop in time?
Part B:
If it were raining instead, how far (in meters) from the intersection would the student need to start braking?
Answer:
18.21 m
41.1 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Part A
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -8.8}\\\Rightarrow s=18.21\ m[/tex]
On a dry day she would have to start braking 18.21 m away from the intersection
Part B
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -3.9}\\\Rightarrow s=41.1\ m[/tex]
On a wet day she would have to start braking 41.1 m away from the intersection
supply 5 N∙m of torque to a bolt located at the origin using a wrench that lies along the positive yy-axis. Assume the bolt head is at the origin and the threaded portion of the bolt lies along the negative xx-axis. If you apply a 25 N force at the end of the wrench in the direction 2???????? − 3????????, what length of wrench would be required to supply the desired torque? Under this force are we tightening or loosening the bolt?
Answer:
Explanation:
We need to supply 5 Nm to the bolt.
If we have a force of 25 N in the direction (2*i - 3*j), the torque is related to the length of the wrench.
T = F * d
d = T / F
d = 5 / 25 = 0.2 m
Assuming the wrench is perpendicular to the force.
The vector of the force is:
[tex]F = 25 N \frac{2i - 3j}{\sqrt{2^2 + (-3)^2}} = 13.9i - 20.8[/tex]
This is a countr-clockwise direction. If the bolt is right handed (most are) we are tightening it.
A special standard weight, known to weight 10 N to high accuracy, is used to check the accuracy of two spring scales. On the first scale the weight gives a reading of 8.4 N. On the second scale the weight gives a reading of 9.1 N. The percent difference between the measurements of the two scales is %. (Never negative.)
Answer:
16 % and 9 %
Explanation:
Accurate reading = 10 N
reading of first balance = 8.4 N
reading of second balance = 9.1 N
Percentage difference in first reading
= [tex]\frac{True reading - reading of first balance}{True reading}\times 100 %[/tex]
= [tex]\frac{10-8.4}{10}\times 100 %[/tex] = 16 %
Percentage difference in second reading
= [tex]\frac{True reading - reading of second balance}{True reading}\times 100 %[/tex]
= [tex]\frac{10-9.1}{10}\times 100 %[/tex] = 9 %
An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) during each cycle. What is the thermal efficiency of the engine?
Answer:
[tex]\eta = 0.411[/tex]
Explanation:
As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine
So here we know that the
input energy = 441.3 kJ
energy rejected = 259.8 kJ
so we have
[tex]Q_1 - Q_2 = W[/tex]
[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]
now efficiency is defined as
[tex]\eta = \frac{W}{Q_1}[/tex]
[tex]\eta = \frac{181.5}{441.3}[/tex]
[tex]\eta = 0.411[/tex]
What is the net charge of a metal ball if there are 21,749 extra electrons in it?
Answer:
[tex]Q=3.47\times 10^{-15}\ C[/tex]
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]
To find the net charge if there are n number of extra electrons is :
Q = n × q
[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]
[tex]Q=3.47\times 10^{-15}\ C[/tex]
So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.
A ball is tossed with enough speed straight up so that it is in the air serveral seconds. Assume upwards direction is positive and downward is negative. What is the change in its velocity during this 1-s interval?
Answer:
- 32.17 fts/s in the imperial system, -9.8 m/s in the SI
Explanation:
We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):
[tex]a = \frac{dv}{dt}[/tex]
So, if we wanna know the change in velocity, we can take the integral:
[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]
Luckily for us, the acceleration in this problem is constant
[tex]a \ = \ - g[/tex]
the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:
[tex]t_f = t_i + 1 s[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = [-g t]^{t_i + 1 s}_{t_i} [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g (t_i+1s) + g t_i [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g t_i + -g 1s + g t_i[/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g 1s [/tex]
taking g in the SI
[tex]g=9.8 \frac{m}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 9.8 \frac{m}{s} [/tex]
or, in imperial units:
[tex]g=32.17 \frac{fts}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 32.17 \frac{fts}{s} [/tex]
Cheetahs can accelerate to a speed of 21.8 m/s in 2.55 s and can continue to accelerate to reach a top speed of 28.1 m/s . Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x direction point in the direction the cheetah runs. Express the cheetah's top speed vtop in miles per hour (mi/h) .
Starting from a crouched position, how much time totall does it take a cheetah to reach its top speed and what distance d does it travel in that time?
If a cheetah sees a rabbit 122.0 m away, how much time ttotal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?
Answer:
Explanation:
Given
Cheetah speed=21.8 m/s in 2.55 sec
i.e. its [tex]a=\frac{21.8}{2.55}=8.55 m/s^2[/tex]
Cheetah top speed=28.1 m/s
And 1 m/s is equal to 2.23694 mph
therefore 28.1 m/s is [tex]2.236\times 28.1=62.858 mph[/tex]
Time taken to reach top speed
v=u+at
[tex]28.1=0+8.55\times t[/tex]
[tex]t=\frac{28.1}{8.55}=3.286 s [/tex]
Distance traveled during this time
[tex]v^2-u^2=2as[/tex]
[tex]28.1^2=2\times 8.55\times s[/tex]
[tex]s=\frac{789.61}{2\times 8.55}=46.176 m[/tex]
If cheetah sees a rabbit 122 m away
time taken to reach rabbit
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]122=0+\frac{1}{2}\times 8.55\times t^2[/tex]
[tex]t^2=\frac{244}{8.55}[/tex]
[tex]t=\sqrt{28.53}=5.34 s[/tex]
During the _____ developmental stage children are drawing their thoughts instead of merely observations.
A. scribbling
B. Schematic
C. circular
D. pre schematic
Answer:
D. pre schematic
Explanation:
During the _pre schematic____ developmental stage children are drawing their thoughts instead of merely observations.
Pre schematic stage in children is the age at which they drawing, symbols, spatial figures to express themselves. Their becomes more complex and full of lines as they age. They use single base lines, multiple base lines and fold up views.
How many photons are emitted per second by a He−Ne laser that emits 1.5 mW of power at a wavelength λ=632.8nm. What is the frequency of the electromagnetic waves emitted by a He−Ne laser?
Answer:
The frequency of the emitted EM wave by He-Ne laser is [tex]4.74\times 10^{14} Hz[/tex]
Given:
Power emitted by He-Ne laser, P = 1.5 mW = [tex]1.5\times 10^{- 3}[/tex]
Wavelength, [tex]\lambda = 632.8 nm = 632.8\times 10^{- 9} m[/tex]
Solution:
Now, to calculate the frequency, [tex]\vartheta[/tex] of the EM wave emitted:
Energy associated with 1 photon = Power of one photon per sec = [tex]\frac{hc}{\lambda}[/tex]
Therefore, power associated with 'N' no. of photons, P = [tex]N\frac{hc}{\lambda}[/tex]
where
[tex]h = 6.626\times 10^{-34} m^{2}kg/s[/tex] = Planck's constant
Now,
[tex]1.5\times 10^{- 3} = N\frac{6.626\times 10^{-34}\times 3\times 10^{8})}{632.8\times 10^{- 9}}[/tex]
N = [tex]4.77\times 10^{15}[/tex]
Also, we know that:
[tex]c = \vartheta \lambda[/tex]
Thus
[tex]\vartheta = \frac{c}{\lambda} = \frac{3\times 10^{8}}{632.8\times 10^{- 9}}[/tex]
[tex]\vartheta = 4.74\times 10^{14} Hz[/tex]
A car traveling 56.0 km/h is 25.0 m from a haystack when the driver slams on the brakes. The car hits the haystack 2.11 s later. How fast is the car traveling at impact?
Answer:
The car is traveling at [tex]8.1366\frac{m}{s}[/tex]
Explanation:
The known variables are the following:
[tex]V_{0} = 56 \frac{km}{h} = 15.56 \frac{m}{s}\\ D=25m\\ t=2.11s\\ V_{f}=?[/tex]
First, from the equation of motion we find the deceleration:
[tex]D=V_{0}*t+\frac{1}{2} a*t^{2} \\ a=\frac{2(D-V_{0})}{t^{2} } \\ a=3.5182\frac{m}{s^2}[/tex]
Then, with the equation for the speed:
[tex]V_{f}=V_{o}+a*t\\ V_{f}=8.1366\frac{m}{s}[/tex]
The speed of the car at impact is approximately 25.72 m/s.
Explanation:To find the speed of the car at impact, we first need to calculate the deceleration. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given that the initial velocity (u) is 56.0 km/h, the distance (s) is 25.0 m, and the time taken (t) is 2.11 s, we can rearrange the equation to solve for a: a = (v - u) / t.
Plugging in the values, we have a = (0 - 56.0 km/h) / 2.11 s = -26.5 m/s². Since the car is decelerating, the acceleration is negative.
Now, to find the final velocity (v) at impact, we can use the equation v² = u² + 2as. Rearranging the equation, we have v² = u² + 2(-26.5 m/s²)(25.0 m) = u² - 2(26.5 m/s²)(25.0 m).
Plugging in the values, we have v² = (56.0 km/h)² - 2(26.5 m/s²)(25.0 m).
Converting the initial velocity (u) to m/s, we get u = 56.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.6 m/s.
Substituting the values, we have v² = (15.6 m/s)² - 2(26.5 m/s²)(25.0 m) = 0.16 m²/s² - 26.5 m²/s² * 25.0 m = 0.16 m²/s² - 662.5 m²/s² = -662.34 m²/s².
Taking the square root of both sides, we get v ≈ √(-662.34 m²/s²) ≈ -25.72 m/s.
Since speed is a positive quantity, the speed of the car at impact is approximately 25.72 m/s.
A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as
[tex]v^2=u^2+2as[/tex]
where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get
[tex]0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s[/tex]
Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as
[tex]v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s[/tex]
Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as
[tex]v=u+at[/tex]
where symbols have the usual meaning
Applying the given values we get
[tex]t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds[/tex]
What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.00 × 10^8 m/s, and 1 nm = 10^−9 m)
Answer:
Wavelength = 736.67 nm
Explanation:
Given
Energy of the photon = 2.70 × 10⁻¹⁹ J
Considering:
[tex]Energy=h\times frequency[/tex]
where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Frequency is:
Frequency = c / Wavelength
So, Formula for energy:
[tex]Energy=h\times \frac {c}{\lambda}[/tex]
Energy = 2.70 × 10⁻¹⁹ J
c = 3×10⁸ m/s
h = 6.63 x 10⁻³⁴ J.s
Thus, applying in the formula:
[tex]2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}[/tex]
Wavelength = 736.67 × 10⁻⁹ m
1 nm = 10⁻⁹ m
So,
Wavelength = 736.67 nm