Answer:
[a]. South direction
(b). 1.354 × 10^14 m/s^2
Explanation:
In order to to be able to solve this question effectively, one must be familiar with physics concepts such as kinetic energy[1/2 ×mass × velocity,v²], magnetic force[ magnetic force = charge × velocity × magnetic field × sin Ф] and Newton's law of motion especially the second law of Newton[ Force = mass × acceleration].
(a). The direction are the electrons deflected by this field component = q( v × B) = - ( i × -k) = -( - ( i × k) = - (- (- j) = -j. The -j is in the south direction.
Note that i × k is being replaced by -j.
(b). In order to determine the magnitude of the acceleration of an electron in part (a), the first thing to do is to calculate the velocity from the equation for kinetic energy.
Thus, velocity = √ [(2 × 2.7 × 10⁻¹⁵) ÷ 9.1 × 10⁻³¹] = 7.7 × 10^7 m/s.
The next thing is to determine the net force.
Therefore, the net force = 1.6 × 10^-19 × 7.7 × 10^7 m/s × 1 × 10^-15 sin 90°.
= 1.232 × 10^ -16N.
Hence, Net Force = mass × acceleration.
1.232 × 10^ -16N = 9.1× 10^-31 × acceleration.
The acceleration = 1.232 × 10^ -16N/ 9.1× 10^-31 = 1.354 × 10^14 m/s^2.
NEED ASAP!! A box of mass 10 kg requires 20 N to slide it across a surface. What is the weight of the box? What is the coefficient of friction between the box and surface ?
Answer: 7
Explanation: 7 is the superior number
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1250 kg when fired, had a muzzle velocity of 750 m/s. Assuming a constant force, determine the explosive force experienced by the shell inside the barrel. Start from a fundamental principle.
Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})[/tex] (1)
Where:
[tex]F[/tex] - Explosive force, measured in newtons.
[tex]\Delta s[/tex] - Barrel length, measured in meters.
[tex]m[/tex] - Mass of the shell, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the shell, measured in meters per second.
If we know that [tex]m = 1250\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 750\,\frac{m}{s}[/tex] and [tex]\Delta s = 15\,m[/tex], then the explosive force experienced by the shell inside the barrel is:
[tex]F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}[/tex]
[tex]F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}[/tex]
[tex]F = 23437500\,N[/tex]
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
An RLC parallel circuit has an applied voltage of 240 volts. R= 60 ohms, XL = 20 ohms, and Xc =36 ohms. What is the capacitor current?
Answer:
6.67A
Explanation:
The voltage across the capacitor formula is expressed as;
VL = IXL
VL is the voltage across the capacitor = 240volts (since it is a parallel connection, all the elements will have the same voltage)
I is the capacitor current
XL is the capacitive reactance = 36 ohms
Recall from the formula:
VL = IXL
I = VL/XL
I = 240/36
I = 6.67A
Hence the capacitor current is 6.67A
Please help with an explanation for 30 points even if u only do one (but with good explanation)
Answer:
1- 4Ω
Explanation:
if 2Ω is 2 meters, than 4 metres is 4Ω
Which property of water causes the cracks in the pavement in cold climates ?
“As temperatures drop, the pavement contracts, building up tensile stresses that lead to cracking,” states MnDOT's Research Services Section. “Fractures occur every 20 to 30 feet across the lane, allowing water to penetrate the structure, which further weakens the pavement layer and the base beneath
Please help me......
can you guys help me to solve first question????
Torque question
A uniform spherical shell of mass M = 11.0 kg and radius R = 0.480 m can rotate about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.160 kg m2 and radius r = 0.110 m, and is attached to a small object of mass m = 1.60 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen a distance 0.700 m after being released from rest? Use energy considerations.
A soccer ball, mass 3 kg, is kicked straight up with a velocity of 25 m/s. How high will the ball travel?
Answer:
h = 31.9 m
Explanation:
When the ball travels from the ground to the highest point, its kinetic energy is converted into potential energy. So by the law of conservation of energy:
[tex]Kinetic\ Energy\ Lost\ by\ ball = Potential\ Energy\ gained\ by\ ball\\\frac{1}{2}mv^{2} = mgh\\\\h = \frac{v^{2}}{2g}[/tex]
where,
h = height gained by ball = ?
v = speed of ball = 25 m/s
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(25\ m/s)^{2}}{2(9.8\ m/s^{2})}[/tex]
h = 31.9 m
Which characterictic of motion could change without changing the velocity of an object
Answer:
The direction could change
What is Nature? How does it explain the human condition?
A boy throws a stone straight upward with an initial
speed of 17.0 m/s.
Part A
What maximum height will the stone reach before falling back down?
Answer:
14.8m
Explanation:
Given parameters:
Initial speed = 17m/s
Unknown:
Maximum height = ?
Solution:
At the maximum height, the final speed will be 0m/s;
We use of the kinematics equation to solve this problem.
V² = U² - 2gH
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height
0² = 17² - (2 x 9.8 x h )
0 = 289 - (9.6h)
-289 = -19.6h
h = 14.8m
Bruce pulls a spring with a spring constant k=100 Nmk=100\, \dfrac{\text N}{\text m}k=100mNk, equals, 100, start fraction, start text, N, end text, divided by, start text, m, end text, end fraction, stretching it from its rest length of 0.20 m0.20\,\text m0.20m0, point, 20, start text, m, end text to 0.40 m0.40\,\text m0.40m0, point, 40, start text, m, end text.What is the elastic potential energy stored in the spring?
Answer:
K_{e} = 2.0 J
Explanation:
In this exercise you are asked to calculate the elastic potential energy of a spring
[tex]K_{e}[/tex] = ½ k x²
where k is the spring constant and x is the displacement from equilibrium position
In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation
Δx = x₁ - x₀
Δx = 0.40 - 0.20
Δx = 0.20 m
let's calculate the elastic potential energy
K_{e} = ½ 100 0.20²
K_{e} = 2.0 J
Which is not a product of coal?
A. Monoxide
B. Carbon Dioxide
C. Methane
D. Calcium
Answer:
D
Explanation:
Hope this helps
A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m. It climbs a frictional incline of height 2.9 m inclined at an angle 16°, then moves along a second frictional surface of coefficient 0.1 before coming to rest.
The acceleration of gravity is 9.8 m/s^2. If the first frictional surface has a coefficient of 0.21 for a distance 1 m, how far does it slide along the second frictional region before coming to rest?
Answer:
[tex]D=99.4665307m \approx 99.5m[/tex]
Explanation:
From the question we are told that
Mass [tex]m=6kg[/tex]
Velocity of mass [tex]V_m=16[/tex]
Force of Tunnel [tex]F_t=8N[/tex]
Length of Tunnel [tex]L_t=1.6[/tex]
Height of frictional incline [tex]H_i=2.9[/tex]
Angle of inclination [tex]\angle =16 \textdegree[/tex]
Acceleration due to gravity [tex]g=9.8m/s^2[/tex]
First Frictional surface has a coefficient [tex]\alpha_1 =0.21\ for\ d_c=1[/tex]
Second Frictional surface has a coefficient [tex]\alpha _2=0.1[/tex]
Generally the initial Kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]K.E=\frac{1}{2}(6)(16)^2[/tex]
[tex]K.E=768[/tex]
Generally the work done by the Tunnel is mathematically given as
[tex]w_t=F_t*d_t[/tex]
[tex]w_t=8*1.6[/tex]
[tex]w_t=12.8J[/tex]
Therefore
[tex]Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t[/tex]
[tex]E_t=K.E+E_t\\E_t=768J+12.8J[/tex]
[tex]E_t=780.8J[/tex]
Generally the energy lost while climbing is mathematically given as
[tex]E_c=mgh[/tex]
[tex]E_c=(6)(9.8)(2.9)[/tex]
[tex]E_c=170.52J[/tex]
Generally the energy lost to friction is mathematically given as
[tex]E_f=\alpha *m*g*cos\textdegree*d_c[/tex]
[tex]E_f=0.21*6*9.8*cos16*1[/tex]
[tex]E_f=11.86965942 \approx 12J[/tex]
Generally the energy left in the form of mass [tex]Em[/tex] is mathematically given as
[tex]E_m=E_t+E_c+E_f[/tex]
[tex]E_m=(768J)-(170.52)-(12)[/tex]
[tex]E_m=585.48J[/tex]
Since
[tex]E_m=\alpha_2*g*m*d[/tex]
Therefore
It slide along the second frictional region
[tex]D=\frac{585.46}{0.1*9.81*6}[/tex]
[tex]D=99.4665307m \approx 99.5m[/tex]
A spring has a spring constant of 200 N/m. How much elastic potential energy is stored in the spring when it is stretched by 0.1 meters? A:2000 J B:20 J C:0 J D:1 J
Answer:
D, 1J
Explanation:
PE=1/2kx^2 and plug in the variables.
PE=1/2 x 200 x 0.1^2= 1J
The distance between a loudspeaker and the left ear of a listener is 2.70 m. (a) Calculate the time required for sound to travel this distance if the air temperature is 20 oc. (b) Assuming that the sound frequency is 523 Hz, how many wavelengths of sound are contained in this distance
Answer:
Explanation:
Distance covered by sound = 2.7 m
speed of sound at 20⁰C = 343 m /s
time take by sound to cover the distance = distance / speed
= 2.7 / 343
= 7.87 ms . ( millisecond )
b )
Wavelength of sound = speed / frequency
= 343 / 523 m
= .6558 m
= 65.58 cm
Distance = 2.70 m = 270 cm
No of wavelengths contained in the distance
= 270 / 65.58
= 4.11 or 4 wavelengths ( by rounding off to digit )
The time taken by sound and the wavelengths of sound is required.
Time taken is [tex]0.00787\ \text{s}[/tex]
The number of wavelenghts is 4.
s = Distance = 2.7 m
v = Speed of sound at [tex]20^{\circ}\text{C}[/tex] = 343 m/s
Time is given by
[tex]t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{2.7}{343}\\\Rightarrow t=0.00787\ \text{s}[/tex]
f = Frequency = 523 Hz
Wavelength is given by
[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{523}\\\Rightarrow \lambda=0.66\ \text{m}[/tex]
The wavelength is 0.66 m.
n = Number of wavelengths
[tex]n\lambda=s\\\Rightarrow n=\dfrac{s}{\lambda}\\\Rightarrow n=\dfrac{2.7}{0.66}\\\Rightarrow n\approx 4[/tex]
The number of wavelenghts is 4.
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A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 193.0 km/h. The conditions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.80 m/s2. What is the minimum required length of the runway
Answer:
513 m
Explanation:
We have;
final speed of the airplane = 193.0 km/h * 1000/3600 = 53.6 m/s
acceleration of the air plane = 2.80 m/s2
initial velocity of the airplane = 0 m/s
length of the runway = distance covered
v^2 = u^2 + 2as
v^2 - u^2 = 2as
s = v^2 - u^2/2a
s = (53.6)^2 - 0^2/ 2 * 2.80
s = 2872.96/ 5.6
s = 513 m
Suppose you have two metal spheres that are exactly the same size, separated by a very large distance. Sphere A carries an excess of eight negative charges while sphere B carries an excess of two positive charges. Sphere A is momentarily connected to Sphere B using a metal wire and then the wire is removed. Draw charge diagrams of the two spheres for the two cases below before and after the connecting wire is used.
1. Before the spheres are connected using the wire .
2. After the wire has been removed .
Answer:
a) phere A has 8 negative charges and sphere B has 2 positive charges
b) each one has then 3 negative charges
Explanation:
In this case, it is asked to determine the charge of the spheres in two conditions
a) Before connecting the cable, sphere A has 8 negative charges and sphere B has 2 positive charges
b) After connecting the cable, as the spheres are metallic, the load is distributed, we have
q = 8 -2 = 6 negative charges
these charges are distributed between the two spheres, each one has then 3 negative charges
Which is the best tool to use when measuring mass
Beam balance is used to determine the mass of an object.
calculated the pressure exerted when a force of 100N is applied on a area of 25m2
Answer:
P = 4[Pa]
Explanation:
Pressure is defined as the relationship between Force over area. So we can use the following equation.
[tex]P=F/A[/tex]
where:
P = pressure [Pa] (units of Pascals)
F = force = 100 [N]
A = area = 25 [m²]
Now replacing:
[tex]P=100/25\\P=4[Pa][/tex]
[tex]\\ \rm\longmapsto P=\dfrac{F}{A}[/tex]
[tex]\\ \rm\longmapsto P=\dfrac{100}{25}[/tex]
[tex]\\ \rm\longmapsto P=4Pa[/tex]
____ and____ are 2 major atmospheric gases
Answer:
Nitrogen and Oxygen are the two major atmospheric gases.
A space station consists of three modules, connected to form an equilateral triangle of side length 82.0 m. Suppose 100 people, with an average mass of 75.0 kg each, live in each capsule and the mass of the modules is negligible compared to the mass of the people. At the current rotational rate the effective acceleration of gravity is g/2. (a) What angular momentum of the system
Answer:
Angular momentum of the system is 16221465.4617 kgm²/s
Explanation:
Given that;
length of the side of the triangle L = 82 M
m = 75.0 kg × 100 = 7500 kg
distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m
effective acceleration a = 9.8 / 2 = 4.9 m/s²
we know that; effective acceleration is being provided by centripetal acceleration.
so
a = R × w²
rate of rotation w = √( a / R) = √( 4.9 / 47.34) = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
we substitute
I = 3 × 7500 × (47.34)²
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
L = 16221465.4617 kgm²/s
Therefore, Angular momentum of the system is 16221465.4617 kgm²/s
The Angular momentum of the system is 16221465.4617 kgm²/s
The calculation is as follows:Given that;
Length of the side of the triangle L = 82 M
[tex]m = 75.0 kg \times 100 = 7500 kg[/tex]
Now
distance of each vertex from center R = L/√3
= 82/√3
= 47.34 m
Now
effective acceleration [tex]a = 9.8 \div 2[/tex] = 4.9 m/s²
So,
a = R × w²
Now
rate of rotation [tex]w = \sqrt( a \div R) = \sqrt( 4.9 \div 47.34)[/tex] = 0.3217 rad/seconds
Moment of Inertia I = 3mR²
Now
[tex]I = 3 \times 7500 \times (47.34)^2[/tex]
Also, Angular momentum L is expressed as;
L = I × w
so
L = 3 × 7500 × (47.34)² × 0.3217
= 16221465.4617 kgm²/s
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34000×0.4
----------------
0.02×200
In scientific notation
Answer:
the simplified expression is written as 3.4 x 10³
Explanation:
Given expression;
[tex]\frac{34000\times 0.4}{0.02 \times 200}[/tex]
in scientific notation, the expression is simplified as;
[tex]\frac{34000\times 0.4}{0.02 \times 200} = \frac{13600}{4} = 3400 = 3.4 \times 10^3[/tex]
Therefore, in scientific notation, the simplified expression is written as 3.4 x 10³
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the charge Q is distributed uniformly, in the shell between r1 and r2, determine the electric field, magnitude and direction, in the following situations:______.
a. From r=0 to r=r1.
b. From r=r1 to r=r2.
c. Outside of r=r2.
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
what is the twisted ladder shape of the DNA called?
Answer:
Double helix
Explanation:
The Double helix is a DNA molecule. The two strands around the Double Helix is called the twisted ladder.
Answer:
double helix
Explanation:
The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000-lb car B is traveling at 6 ft/s to the left. If the cars crash head-on, and at a time instant during the crash the impact force on A is 900 lb to the left, what is magnitude and direction of the impact force exerted on B at the instant
Answer:
900 lb to the right
Explanation:
Newton's third law of motion states that for every action there is equal and opposite reaction.
Hence if the two cars (i.e. car A and car B) crash head on and there is an impact on car A of 900 lb to the left, car B would have to generate a force of equal magnitude and in an opposite direction. Since car A had a force of 900 lb to the left, the impact force exerted on car B would be 900 lb to the right.
The answer is 900 lb to the right
A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0260 kg, and its speed is 42.8 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = [tex]p_{fx} - p_{ox}[/tex]
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m [tex]v_{fx}[/tex] - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
[tex]\Delta p_{y}[/tex] = 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
Solve ez points. I cant rlly get but i guess most of u do, so its ez
Answer:
A: nucleus
D: neutron
hope this helps :D
Answer:
A: nucleus
D: neutron
Explanation: hope this helps! <3
The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.4 m/s in the positive x direction and some time later has a velocity of 6.4 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
33.5J
Explanation:
Given:
Mass of the canister= 3.1 kg
Initial velocity of canister= v(i)= 4.4i m/s
Final velocity of canister= v(f)= 6.4j m/s
Force magnitude( xy plane)= 5 N
The magnitude of vector V'= Vxi + Vyj + Vzk
|V|= √( Vx^2 + Vy^2 + Vz^2
From Kinectic energy and work theorem.
Net work = Kinectic energy of the canister
ΔK= W
(Kf - Ki)= W
Where Kf= final Kinectic energy
= 1/2 mv^2
If we input the given values we have,
= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J
Ki= initial Kinectic energy
= 1/2 mv^2
If we substitute the given values we have
=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J
Work done by canister = (final Kinectic energy - initial energy)
= 63.5- 30
=33.5J
Hence, work done on the canister 33.5J
List two things affected by friction PLEASE RESPOND I NEED HELP
metal and hair I think hope this helps
Answer:
cars and trains
Explanation:
i hope that good
