Answer:
(a) 0.3 C
(b) 0.9 J
Explanation:
(a)
Given:
Current drawn (I) = 0.50 mA = 0.50 × 10⁻³ A
Terminal voltage (V) = 3.0 V
Time of operation (t) = 10 min = 10 × 60 = 600 s
Charge flowing through the toy car 'Q' is given as:
[tex]Q=It[/tex]
Plug in the given values and solve for 'Q'. This gives,
[tex]Q=(0.50\times 10^{-3}\ A)(600\ s)\\\\Q=0.3\ C[/tex]
Therefore, 0.3 C charge flows the toy car.
(b)
Energy lost by the battery is equal to the product of power consumed by the battery and time of operation.
Power consumed by the battery is given as:
[tex]P=VI[/tex]
Plug in the given values and solve for 'P'. This gives,
[tex]P=(3.0\ V)(0.50\times 10^{-3}\ A)\\\\P=1.5\times 10^{-3}\ W[/tex]
Therefore, the energy lost by the battery is given as:
[tex]E=P\times t\\\\E=1.5\times 10^{-3}\ W\times 600\ s\\\\E=0.9\ J[/tex]
Therefore, the energy lost by the battery is 0.9 J
A backcountry skier weighing 700 N skis down a steep slope, unknowingly crossing a snow bridge that spans a deep, hidden crevasse. The bridge can support 470 N − meaning that's the maximum normal force it can sustain without collapsing.
Answer:
The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°
Explanation:
Given that,
Weight = 700 N
Normal force = 470 N
Suppose, Find the minimum slope angle for which the skier can safely traverse the snow bridge.
We need to calculate the minimum slope angle
Using balance equation
[tex]F_{N}=mg\cos\theta[/tex]
[tex]\theta=\cos^{-1}(\dfrac{F_{N}}{mg})[/tex]
Put the value into the formula
[tex]\theta=\cos^{-1}(\dfrac{470}{700})[/tex]
[tex]\theta=47.82^{\circ}[/tex]
Hence, The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°
When a 700 N skier crosses a snow bridge that can only support 470 N, the bridge will collapse due to the skier's weight exceeding the bridge's maximum normal force. Newton's second law can be used to understand this situation, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The relationship between weight, mass, and the gravitational force can also be used to determine that the normal force should be equal to the skier's weight for the bridge to be safe.
Explanation:The maximum normal force that the snow bridge can sustain is 470 N. The backcountry skier weighs 700 N, so the bridge will collapse under their weight.
To understand why, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The skier's weight is acting downwards, while the normal force of the snow bridge is acting upwards. When the skier crosses the bridge, the normal force should be equal to their weight for the bridge to be in equilibrium. However, since the normal force is less than the skier's weight, the bridge cannot sustain the weight and will collapse.
This situation can be described using the formula: Weight = mass x gravitational force. So, we can calculate the mass of the skier by dividing their weight by the gravitational force. With a weight of 700 N and a gravitational force of 9.8 m/s², the mass is approximately 71.4 kg. Therefore, the normal force supporting the skier on the bridge should be equal to 700 N for it to be safe.
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Four forces of the same magnitude but differing directions act at and tangent to the rim of a uniform wheel free to spin about its center of mass (CM). Which statement about the wheel is correct?
A. None of these
B. Neither the net force on the wheel nor the net torque on the wheel about the CM
is zero.
C. The net force on the wheel is not zero but the net torque about the CM is zero.
D. The net force on the wheel and the net torque on the wheel about the CM are zero.
E. The net force on the wheel is zero but the net torque about the CM is not zero.
Final answer:
The correct statement about a wheel with four forces of the same magnitude but different directions acting tangentially at its rim is that both the net force on the wheel and the net torque about the CM are zero.
Explanation:
When four forces of the same magnitude but differing directions act tangentially at the rim of a uniform wheel free to spin about its center, the net force acting on the wheel may be zero if the forces cancel each other out. However, the direction and point of application of these forces are critical in determining the net torque. If the forces are applied in such a way that they create rotational effects that cancel each other, the net torque about the center of mass (CM) will also be zero. Therefore, the correct statement about the wheel is:
D. The net force on the wheel and the net torque on the wheel about the CM are zero.
To put it simply, the forces are arranged in a way that causes them to balance out, leading to no linear acceleration (net force is zero), and they are positioned symmetrically around the center of mass, producing no rotational acceleration (net torque is zero).
While describing a circular orbit 300 mi above the earth a space vehicle launches a 6000-lb communications satellite. Determine the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 22,000 mi
Answer:
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
Explanation:
While the satellite is in the space vehicle, it has the next potential energy
[tex]U = -\frac{GmMe}{r}[/tex]
where G is the gravitational constant
m is the satellite's mass in kilograms
Me is the earth's mass
r is the orbit's radius from to the earth's center in meters
[tex]U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}[/tex]
[tex]U = -2.2423x10^{12} J[/tex]
The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi
[tex]U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}[/tex]
[tex]U = -29696124610.3 J[/tex]
Then
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
The specific surface energy for aluminum oxide is 0.90 N/m, and the elastic modulus is 393 GPa. Compute the critical stress, in MPa, required for propagation of a surface crack of length 0.25 mm. Round answer to 3 significant figures and report in the format: 12.3 MPa
Answer:
42.4 Npa
Explanation:
Explanation is attached in the picture below
A human being can be electrocuted if a current as small as 55 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding. If his resistance is 2300 Ω, what might the fatal voltage in volts be?
Answer:
The lethal voltage for the electrician under those conditions is 126.5 V.
Explanation:
To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:
V = i*R
Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:
V = 0.055*2300 = 126.5 V.
The lethal voltage for the electrician under those conditions is 126.5 V.
Answer:
126.5 V
Explanation:
Using Ohm's Law,
V = IR............................. Equation 1
Where V = Voltage, I = current, R = resistance.
Note: The current needed to bring about the fatal voltage is equal to the current that will cause human being to be electrocuted.
Given: I = 55 mA = 55/1000 = 0.055 A, R = 2300 Ω
Substitute into equation 1
V = 0.055×2300
V = 126.5 V.
Hence the fatal voltage = 126.5 V
A phone cord is 4.67 m long. The cord has a mass of 0.192 kg. A transverse wave pulse is produced by plucking one end of the taunt cord. The pulse makes four trips down and back along the cord in 0.794 s.
What is the tension in the cord?
Answer:
91.017N
Explanation:
Parameters
L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)
so N= no of trips = 8, From Wave speed(V) = N *L/t , we have :
V= 8*4.67/0.794 = 47.0529 m/s.
We compute the cords mass per length, Let it be P
P = M/L = 0.192/4.67 = 0.04111 kg/m
From T = P * V^2 where T = Tension, we have
T = 0.04111 * (47.0529)^2
T = 91.017N.
The tension in the cord is 91.017N
A 925 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?
Answer:
[tex]\mu_s^{min}=0.885[/tex]
Explanation:
The centripetal force is provided by the static friction between the car and the road, and always have to comply with [tex]f\leq\mu_sN[/tex], so we have:
[tex]ma_{cp}=f\leq\mu_sN=\mu_smg[/tex]
Which means:
[tex]a_{cp}=\frac{v^2}{r}\leq\mu_sg[/tex]
So we have:
[tex]\frac{v^2}{gr}\leq\mu_s[/tex]
Which means that [tex]\frac{v^2}{gr}[/tex] is the minimum value the coefficient of static friction can have, which for our values is:
[tex]\mu_s^{min}=\frac{v^2}{gr}=\frac{(25m/s)^2}{(9.81m/s^2)(72m)}=0.885[/tex]
To keep the car from skidding, the frictional force between the car and the road must provide the necessary centripetal force to keep the car moving in a circle. The minimum coefficient of friction required would be about 0.86.
Explanation:To determine this, we must understand that the friction force between the car and the road surface is what keeps the car in a curved path. If the speed of the car or the radius of the curve is too large, a greater force is needed to keep the car from skidding. This force must be supplied as an increased friction force, which implies a larger coefficient of friction.
The centripetal force needed to keep a car on the road as it rounds a turn is provided by the frictional force between the road and the car's tires. We can use the formula for centripetal force where Fc=m*v^2/r. Here, Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the curve. In this case, friction provides the centripetal force, so Fc is also equal to the force of static friction, which is less than or equal to the coefficient of static friction times the normal force (µ*m*g).
Setting these equal and solving for µ gives µ=v^2/(g*r). Plugging in the given values (v=25 m/s, g=9.8 m^2/s, and r=72 m), we find that the minimum coefficient of friction required is about 0.86. This is reasonable; a typical car with good tires on dry concrete requires a minimum coefficient of friction of about 0.7 to keep from skidding in a curve.
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The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch high in a tree. Fortunately, they have a spring they can use to launch a rock. The spring constant is 1000 N/m, and they can compress the spring a maximum of 30 cm. All the rocks on the island seem to have a mass of 400 g.a. With what speed does the rock leave the spring?b. If the fruit hangs 15 m above the ground, will they feast or go hungry?
Answer:
(a) v = 15m/a
(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.
Explanation:
Please see the attachment below for film solution.
The rock does not reach the fruit because the maximum height it achieves is 11.48 meters, which is below the required 15 meters. Therefore, the contestants will go hungry.
Let's solve the problem step by step. First, we'll determine the speed with which the rock leaves the spring, and then we'll check if this speed is enough to reach the fruit hanging in the tree.
Part (a): Speed of the rock leaving the spring
Using the principle of conservation of energy, the potential energy stored in the spring when compressed will be converted into the kinetic energy of the rock when the spring is released.
The potential energy stored in the spring [tex]\( E_{\text{spring}} \)[/tex] is given by:
[tex]\[E_{\text{spring}} = \frac{1}{2} k x^2\][/tex]
Substituting the given values:
[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \, \text{N/m} \times (0.30 \, \text{m})^2\][/tex]
[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \times 0.09\][/tex]
[tex]\[E_{\text{spring}} = 45 \, \text{J}\][/tex]
This energy will be converted into kinetic energy [tex]\( E_{\text{kinetic}} \)[/tex] of the rock:
[tex]\[E_{\text{kinetic}} = \frac{1}{2} m v^2\][/tex]
Equating the spring potential energy to the kinetic energy:
[tex]\[45 \, \text{J} = \frac{1}{2} \times 0.4 \, \text{kg} \times v^2\][/tex]
Solving for ( v ):
[tex]\[45 = 0.2 \times v^2\][/tex]
[tex]\[v^2 = \frac{45}{0.2}\][/tex]
[tex]\[v^2 = 225\][/tex]
[tex]\[v = \sqrt{225}\][/tex]
[tex]\[v = 15 \, \text{m/s}\][/tex]
So, the speed of the rock as it leaves the spring is [tex]\( 15 \, \text{m/s} \).[/tex]
Part (b): Checking if the rock can reach the fruit
To determine if the rock can reach the height of 15 m, we use the following kinematic equation for vertical motion:
[tex]\[h = v_0 t - \frac{1}{2} g t^2\][/tex]
First, determine the time ( t ) it takes to reach the maximum height where the vertical velocity becomes zero:
[tex]\[v = v_0 - g t\][/tex]
At the maximum height, [tex]\( v = 0 \):[/tex]
[tex]\[0 = 15 - 9.81 t\][/tex]
[tex]\[t = \frac{15}{9.81}\][/tex]
[tex]\[t \approx 1.53 \, \text{s}\][/tex]
Now, calculate the maximum height [tex]\( h_{\text{max}} \):[/tex]
[tex]\[h_{\text{max}} = v_0 t - \frac{1}{2} g t^2\][/tex]
[tex]\[h_{\text{max}} = 15 \times 1.53 - \frac{1}{2} \times 9.81 \times (1.53)^2\][/tex]
[tex]\[h_{\text{max}} = 22.95 - 11.47\][/tex]
[tex]\[h_{\text{max}} = 11.48 \, \text{m}\][/tex]
The maximum height reached by the rock is approximately 11.48 m, which is less than the 15 m needed to reach the fruit.
A laboratory dish, 20 cm in diameter, is half filled with water. One at a time, 0.49 μL drops of oil from a micropipette are dropped onto the surface of the water, where they spread out into a uniform thin film. After the first drop is added, the intensity of 640 nm light reflected from the surface is very low. As more drops are added, the reflected intensity increases, then decreases again to a minimum after a total of 13 drops have been added. What is the index of refraction of the oil?
Explanation:
Formula for path difference is as follows.
x = 2tn
and, refractive index (n) = [tex]\frac{\lambda}{2t}[/tex]
Thickness is calculated as follows.
Thickness (t) = [tex]\frac{volume}{area}[/tex]
Area = [tex]\pi r^{2}[/tex]
= [tex]\pi \times (\frac{d}{2})^{2}[/tex]
= [tex]\frac{0.49 \times 10^{-6}}{3.14 \times 0.01 m}[/tex]
= [tex]1.56 \times 10^{-8}[/tex] m
Now, the refractive index will be calculated as follows.
For drop, n = [tex]\frac{\lambda}{2t}[/tex]
For B drop, n = [tex]\frac{\lambda}{26t}[/tex]
So, n = [tex]\frac{640 \times 10^{-9}}{26 \times 1.56 \times 10^{-8}}[/tex]
= [tex]\frac{640 \times 10^{-9}}{40.56 \times 10^{-8}}[/tex]
= 1.5
Thus, we can conclude that index of refraction of the oil is 1.5.
If the potential in a region is given by the function V = 2 x − y 2 − cos(z), what is the y-component of the electric field at the point P = (x ′ , y ′ , z ′ )?
Answer:
2y
Explanation:
Electric field in terms of Electric potential is given as:
E = dV/dr(x, y, z)
Where r(x, y, z) = position in x, y, z plane
The y component of the Electric field will be:
Ey = -dV/dy
Given that
V = 2x - y² - cos(z)
dV/dy = -2y
=> E = - (-2y)
E = 2y
A 640 kg automobile slides across an icy street at a speed of 63.9 km/h and collides with a parked car which has a mass of 816 kg. The two cars lock up and slide together. What is the speed of the two cars just after they collide?
Answer:
V3 = 7.802 m/s
Explanation:
m1 = 640 Kg, M2 = 816 kg, V1 = 63.9 Km/h = 17.75 m/s, V2 =0 m/s
Let V3 is the combine velocity after collision.
According to the law of conservation of momentum
m1 v1 + m2 v2 = (m1 + m2) v3
⇒ V3 =( m1 v1 + m2 v2 ) / (m1 + M2)
V3 = ( 640 Kg × 17.75 m/s + 816 kg × 0m/s) / (640 Kg + 816 kg)
V3 = 7.802 m/s
Answer:
28.088 km/h or 7.802 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')....................... Equation 1
Where m = mass of the automobile, m' = mass of the car, u = initial velocity of the car, u' = initial velocity of the car, V = velocity of the two car after collision
Make V the the subject of the equation
V = (mu+m'u')/(m+m')................ Equation 2
Given: m = 640 kg, m' = 816 kg, u = 63.9 km/h, u' = 0 m/s (parked).
Substitute into equation 2
V = (640×63.9+816×0)/(640+816)
V = 40896/1456
V = 28.088 km/h = 28.088(1000/3600) m/s = 7.802 m/s
Hence the speed of the two cars after they collide = 28.088 km/h or 7.802 m/s
Problem 14.45 Two small spheres A and B, of mass 2.5 kg and 1 kg, respectively, are connected by a rigid rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity Determine 0 v (3.5 m/s) .i (a) the linear momentum of the system and its angular momentum about its mass center G, (b) the velocities of A and B after the rod AB has rotated through 180 .
Answer:
(a) linear momentum = 8.75. angular momentum =0. (b) 2.5m/s
Explanation:
A and B are on horizontal and so is the velocity given to A = 3.5m/s i (in i direction which is x direction ), this means that there would be motion only in x direction.
let's first calculate linear momentum.
(A)
p(total)= P_A+P_B = (2.5Kgx3.5m/s)+(1kgx0m/s) = 8.75m/s.
Angular momentum.
L=m*v x r. as the motion is only in x direction and there is no rotation in the system, there fore m*v x r = 0
(B)
Velocity of A and B come from the fact that total linear momentum is conserved.
p_final = P_A+P_B = (mA=mB)v_new = mA*V1+mB*vB.
second term on right is = 0 because B has 0 velocity.
solving for V_new and with the values of all unknown substituted in gives
V_new = (mA*VA)/(mA+mB)= 8.75/3.5= 2.5m/s
The diagram showing the 2 spheres is missing, so i have attached it.
Answer:
A) Linear momentum =(8.75 kg/m.s)i
Angular momentum = (-0.5kg.m²/s)k
B) Va' = (1.5m/s)i and Vb' = (5 m/s)j
Explanation:
First of all, let's find the position of the mass centre;
y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m
A) Linear momentum is given as;
L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i
Angular momentum is given as;
HG = Vector GA x m(a) x v(o)
Where vector GA is the position of the mass centre;
Thus;
HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k
B) from conservation of linear momentum;
MaVo = Ma Va' + Mb Vb
2.5 x 3.5 = 2.5Va' + 1 Vb'
8.75 = 2.5Va' + 1 Vb' - - - - eq(1)
Also, for conservation of angular momentum ;
raMaVo = - raMa Va' + rbMb Vb'
from the diagram attached, ra + rb = 0.2
Now, ra is the same as value as that of the centre of the mass.
Thus, ra = 0.057143.
rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286
Thus;
0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'
0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2
Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s
(e) Imagine the Moon, with 27.3% of the radius of the Earth, had a charge 27.3% as large, with the same sign. Find the electric force the Earth would then exert on the Moon.
Explanation:
Below is an attachment containing the solution.
To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
Explanation:To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electric force is:
F = k * (q1 * q2) / r^2
Where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the charge of the Moon is 27.3% of the charge of the Earth, and the radius of the Moon is also 27.3% of the radius of the Earth. Using these values, we can calculate the electric force.
Let's assume the charge of the Earth is q1 and the charge of the Moon is q2. Since the charge of the Moon is 27.3% as large as the charge of the Earth, we can write q2 = 0.273 * q1. Similarly, the radius of the Moon is 27.3% of the radius of the Earth, so we can write r = 0.273 * R, where R is the radius of the Earth. Plugging these values into Coulomb's law formula:
F = k * (q1 * (0.273 * q1)) / (0.273 * R)^2
Simplifying the equation, we get:
F = k * (q1^2 * 0.273) / (0.273^2 * R^2)
The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
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Consider a rotating object. On that object, select a single point that rotates. How does the angular velocity vector of that point compare to the linear velocity vector at that point? a. perpendicular b. parallel c. neither parallel or perpendicular d. Depends on the rotation
Answer:
a. Perpendicular.
Explanation:
The relation between angular velocity vector and linear velocity vector is described by a cross product:
[tex]\vec v = \vec \omega \times \vec r[/tex]
Where [tex]\vec r[/tex] is perpendicular to the rotation axis and [tex]\vec \omega[/tex] is parallel to the same axis. By definition of cross product, [tex]\vec v[/tex] is a vector which is perpendicular to both vectors. Therefore, linear velocity vector is perpendicular to angular velocity vector. The correct answer is A.
The angular velocity vector, which points along the rotation axis, is perpendicular (Option A) to the linear velocity vector, which is tangent to the rotation path.
Explanation:The angular velocity of a point on a rotating object is a vector that points along the axis of rotation. This vector's direction is determined with respect to the right-hand rule: if the fingers on your right-hand curl from the x-axis toward the y-axis, your thumb points in the direction of the positive z-axis. For an angular velocity pointing along the positive z-axis, the rotation is counterclockwise, whereas for an angular velocity pointing along the negative z-axis, the rotation is clockwise.
On the other hand, the linear velocity of the same point is a vector that is tangent to the path of rotation. It represents the instantaneous linear speed of the point as it moves along the path.
Therefore, given the definitions and directions of both vectors, the angular velocity vector of a point on a rotating object is perpendicular to the linear velocity vector of that point.
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A Global Positioning System (GPS) functions by determining the travel times for EM waves from various satellites to a land-based GPS receiver. If the receiver is to detect a change in travel distance of the order of 3 m, what is the associated travel time in (in ns) that must be measured
Answer:
10ns
Explanation:
Suppose the EM wave travels at light speed [tex]c = 3\times10^8 m/s[/tex]. A change in travel distance of the order of 3 m would result of a change in travel time of
[tex] \Delta t = \frac{\Delta s}{c} = \frac{3}{3\times10^8} = 10^{-8} s = 10 ns[/tex]
To detect a change in travel distance of about 3 m, a GPS receiver must measure a change in travel time of approximately 10 nanoseconds. This is calculated using the speed of light and the time = distance/speed formula.
Explanation:To determine the associated travel time that must be measured for a change in travel distance of 3 meters, we need to know the speed of electromagnetic waves, which is the speed of light, typically denoted 'c'. The speed of light is approximately 3.00 x 108 meters per second (m/s).
Now we can calculate the time it takes for light to travel a given distance using the formula time (t) = distance (d) / speed (s). Substituting for our given distance (3 m) and the speed of light, we get t = 3 m / 3.00 x 108 m/s. This results in a time of approximately 1 x 10-8 seconds or 10 nanoseconds (ns).
Therefore, to detect a change in travel distance of about 3 m, the GPS receiver must be able to measure a change in travel time of roughly 10 nanoseconds.
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Suppose there are single-particle energy eigenvalues of 0, e, 2.::, and 3.:: which are non-degenerate. A total of 6.0 is to be shared between four particles. List the configuration of the particles and their degeneracies for: distinguishable particles; indistinguish3:ble Bose particles; indistinguishable Fermi particles
Answer:
so 40 ways for distinguishable
4 ways for bosons
1 way for fermions
Explanation:
The explanation is attached below
At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If qqq is positive, what is the direction of the force on the particle due to the magnetic field?
Answer:
Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction
Explanation:
The equation for the force exerted on a particle due to the magnetic field is equal to:
F = q*(v * B)
If we replace vi for v, +q for q and -Bk for B, we have:
F = +q*(vi*(-Bk)) = -q*v*B*(i * k)
From this equation we have that the vector i is in direction -x and k in direction -z
We compute the cross product of two unit vectors:
i * k = -j, where j is the vector along -y direction
Replacing we have:
F = -q*v*B*(-j) = -q*v*B*j
Therefore, the force exerted by the magnetic field on the particle goes along the positive y- direction
When driving straight down the highway at a constant velocity you have to give the engine a little gas (which means an added external force of tires pushing on the road) to maintain your uniform motion. Explain why this does not violate Newton’s 1st law. What are the other forces acting on the vehicle?"
Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.
Applying extra gas to a car moving at a constant velocity does not violate Newton's 1st Law as the additional force is required to counteract other forces, such as air resistance and road friction, allowing the car to maintain its velocity. This is, in fact, a confirmation of Newton's 1st Law.
Even while driving at a constant velocity on a straight highway, one must occasionally apply a bit of extra gas, which means applying an extra external force.
However, this does not violate Newton's 1st Law, or the law of inertia, which states that an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
The reason for this is the existence of other forces acting on the car that work against the forward motion. These can include air resistance (or drag), friction from the road, and uphill forces if the road is inclined.
The extra force (gas) provided helps overcome these forces, allowing the car to maintain a constant velocity. So, this activity is actually a confirmation of Newton's first Law, not a violation.
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The 2013 Toyota Camry has an empty weight of 3190 lbf, a frontal area of 22.06 ft2 , and a drag coeffi cient of 0.28. Its rolling resistance is Crr < 0.035. Estimate the maximum velocity, in mi/h, this car
Answer:
V = 144mi/h
Explanation:
Please see the attachment below.
calorimeter has aluminum inner cup of mass 120 gram containing 100 ml water at temperature 20 degree Celsius. Brass piece with mass 100 gram is heated to 100 degree Celsius, and then immersed in the calorimeter. Calculate the final temperature of the system. The specific heat of brass is 0.09 cal/(gramXdeg.C) . The additional necessary data are provided in the text.
Answer:
the final temperature of the system of the system is 25.32°C
Explanation:
We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water
The aluminium calorimeter has a mass Mc= 120g
Volume of water in calorimeter = 100ml at θc =20°C
Density of water is
1000Kg/m³ = 1g/mL
Then, density = mass/ volume
Mass=density ×volume
Mass=1g/mL×100mL
Mass=100gram
Then, the mass of water is
Mw = 100gram
Mass of brass is Mb = 100gram
The temperature of brass is θb=100°C
The specific heat capacity of water is Cw= 1cal/g°C
The specific heat capacity of aluminum Ca=0.22cal/g°C
We are looking for final temperature θf=?
Given that the specific heat capacity of brass is Cb=0.09Cal/g°C
Using the principle of calorimeter;
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, the calorimeter gained heat and the liquid in the calorimeter gain heat too
Heat gain by calorimeter(Hc) = Mc•Ca•∆θ
Where Mc is mass of calorimeter,
Ca is Specific Heat capacity of Calorimeter
∆θ=(θf-θc)
Hc=Mc•Ca•∆θ
Hc=120•0.22•(θf-20)
Hc=26.4(θf-20)
Hc=26.4θf-528
Also, heat gain by the water
Heat gain by wayer(Hw) = Mw•Cw•∆θ
Where Mw is mass of water,
Cw is Specific Heat capacity of water
∆θ=(θf-θw),
Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C
Hw=Mw•Cw•∆θ
Hw=100•1•(θf-20)
Hw=100(θf-20)
Hw=100θf-2000
Also heat loss by the brass is given by
heat loss by brass
Heat loss by brass(Hb)= Mb•Cb•∆θ
Where Mb is mass of brass,
Cb is Specific Heat capacity of brass
∆θ=(θb-θf)
Therefore,
Hb=Mb•Cb•∆θ
Hb=100•0.09•(100-θf)
Hb=9(100-θf)
Hb=900-9θf
Applying the principle of calorimeter
Heat gain = Heat loss
Hc+Hw=Hb
26.4θf-528 + 100θf-2000=900-9θf
26.4θf+100θf+9θf=900+2000+528
135.4θf=3428
Then, θf=3428/133.4
θf=25.32°C
The final temperature of the system is 25.32 degree Celsius.
Given data:
The mass of aluminum cup is, m = 120 g .
The mass of brass piece is, m' = 100 g.
The volume of water in aluminum cup is, V = 100 ml.
The temperature of water is, T = 20 degree Celsius.
The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].
The temperature of brass is, T'' = 100 degree Celsius.
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, first calculate heat gain by calorimeter,
H = mc (T' - T)
Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,
[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]
And, heat gain by water is,
[tex]H'=m'c'(T'-T)[/tex]
Here, c' is the specific heat of water. ( c' =1 )
Solving as,
[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]
Now, heat loss by brass is,
[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]
Applying the principle of calorimeter
Heat gain = Heat loss
H + H' = H''
(26.4T' - 528) + (100T' - 2000) = (900 - 9T')
135.4 T' = 3428
T ' = 25.32 degree Celsius
Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.
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a heavy box is pulled across the floor with a rope. The rope makes an angle of 60 degrees with the floor. A force of 75 N is exerted on the rope. What is the component of the force parallel to the floor
Answer:
37.5 N
Explanation:
Horizontal component is represented by Fx and is given as
Fx = F cos θ
Here,
θ = 60 degrees
F= 75 N
So,
Fx= 75 cos (60°)
==> Fx = 75 × 0.5 =37.5 N
Final answer:
The horizontal component of the force acting parallel to the floor when a 150 N force is applied at a 60-degree angle to the horizontal is 75 N, calculated using the cosine of the angle.
Explanation:
The question involves resolving a force into its components, which is a common problem in physics. When a force is applied at an angle to the horizontal, it has both horizontal and vertical components. The horizontal (parallel) component ([tex]F_{parallel}[/tex]) is found using the cosine function of the angle Θ , which in this case is 60 degrees. The formula is [tex]F_{parallel}[/tex] = F * cos(Θ), where F is the magnitude of the force.
Given that a force of 150 N is applied at an angle of 60 degrees to the horizontal, the horizontal component of the force can be calculated as follows:
[tex]F_{parallel}[/tex] = 150 N * cos(60°)
[tex]F_{parallel}[/tex] = 150 N * 0.5
[tex]F_{parallel}[/tex] = 75 N
The horizontal component of the force acting parallel to the floor is 75 N. To find the net force acting on the box and thus the acceleration, you would subtract the frictional force from the parallel component of the applied force and then use Newton's second law, F = m * a, where F is the net force, m is the mass of the object and a is the acceleration. However, the frictional force is not needed for this particular question about the horizontal component.
Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
58.6 N
Explanation:
We are given that
[tex]q_1=-7.97\mu C=-7.97\times 10^{-6} C[/tex]
[tex]q_2=3.55\mu C=3.55\times 10^{-6} C[/tex]
Using [tex]1\mu C=10^{-6} C[/tex]
[tex]r=6.59 cm=6.59\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
The magnitude of force that one particle exerts on the other
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=9\times 10^9[/tex]
Substitute the values
[tex]F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}[/tex]
F=58.6 N
Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizontally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?
Answer and Explanation:
The answer is attached below
The velocity of both the objects after the collision is [tex]\frac{25}{11} m/s[/tex].
(a) the relative velocity of the masses is zero.
(b) the compression of the spring is 0.185m.
Inelastic collision:
The given case is an example of an inelastic collision in which two objects after the collision move together. The momentum of the system is conserved, therefore,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
here, m₁ = 5kg , m₂ = 6kg
u₁ = 5 m/s , initial velocity of mass m₁, and
u₂= 0, initial speed of the mass m₂
[tex]5\times5+0=(5+6)v\\\\v=\frac{25}{11}m/s[/tex]
(a) When the spring is fully compressed both the masses move with the same velocity, therefore the relative speed of the masses is zero.
(b) from the law of conservation of energy:
the initial kinetic energy of the masses is converted into final kinetic energies and the potential energy of the spring:
[tex]\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u^2_2=\frac{1}{2} (m_1+m_2)v^2+\frac{1}{2}kx^2[/tex]
where x is the compression of the spring
[tex]\frac{1}{2}\times5\times5^2+0=\frac{1}{2}(5+6)\times(\frac{25}{11})^2=\frac{1}{2}\times2000x^2\\\\x=0.185m[/tex]
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Be sure to answer all parts. To improve conductivity in the electroplating of automobile bumpers, a thin coating of copper separates the steel from a heavy coating of chromium. (a) What mass of Cu is deposited on an automobile trim piece if plating continues for 1.25 h at a current of 5.1 A?
Answer:
Explanation:
Total charge passed
= 1.25 x 60 x 60 x 5.1 C
= 22950 C
Equivalent mass of copper
= 63.5 / 2
= 31.75 g
96500 coulomb is required to obtain 31.75 g of copper
22950 C will release (31.75 / 96500) x 22950
= 7.55 g of copper .
Ask Your Teacher Given the displacement vectors A with arrow = (5.00 î − 7.00 ĵ + 5.00 k) m and B with arrow = (3.00 î + 7.00 ĵ − 3.00 k) m, find the magnitudes of the following vectors and express each in terms of its rectangular components.
Answer: [tex]||\vec A|| \approx 9.95 m,||\vec B|| \approx 8.19 m[/tex]
Explanation:
Magnitudes can be calculated by using Pythagorean theorem:
[tex]||\vec A|| =\sqrt{(5m)^{2}+(-7m)^{2}+(5m)^{2}}\\||\vec A|| \approx 9.95 m\\||\vec B|| =\sqrt{(3m)^{2}+(7m)^{2}+(-3m)^{2}}\\||\vec B|| \approx 8.19 m\\[/tex]
Suppose a person riding on top of a freight car shines a searchlight beam in the direction in which the train is traveling. Compare the speed of the light beam relative to the ground when:
a. The train is at rest.
b. The train is moving.
How does the behavior of the light beam differ from the behavior of a bullet fired in the same direction from the top of the freight car?
Answer:
Explanation:
a ) The speed of light will be 3 x 10⁸ m /s
b ) Speed of light will again be 3 x 10⁸ m /s . for an observer on the ground. It is so because speed of light is independent of moving frame of reference . It is absolute . It can not be changed by changing frame of reference .
The behavior of light differs because the speed of bullet will be less in case of stationary train . For moving train , speed of bullet will be increased by amount equal to speed of train for an observer on the ground.
An object of mass m is lowered at constant velocity at the end of a string of negligible mass. As it is lowered a vertical distance h, its gravitational potential energy changes by ∆Ug = −m g h. However, its kinetic energy remains constant, so that if we define E = K + Ug, we find ∆E = −m g h. Why isn’t the total energy E conserved? 1. Because the universe is accelerating in its expansion, the object is actually at rest and not descending ... the earth moves away as fast as it moves "down." 2. An external force is doing work on the system. 3. In reality, all objects are massless, so that m = 0 and ∆E = 0. 4. The acceleration of the system is zero. 5. The net force on the system is not zero. 6. Ug is defined incorrectly as if gravity were a constant force. 7. The total energy is indeed conserved, since ∆E = ∆Ug. 8. E is useless in real-world examples like this.
Answer:
Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.
The conserved form of the equation is given by
E = K + Ug + Wother.
In this case Wother = work done by the tension force.
In that form the total mechanical energy is conserved.
Final answer:
The correct answer is: An external force is doing work on the system.
Explanation:
When the object is lowered at constant velocity, an external force (in this case, the force exerted by the person lowering the object) is doing work against gravity to keep the velocity constant. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh).
Since work is being done on the system externally, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved.
However, in the scenario described, an external force is doing work on the system. Work is being done to counteract gravity and maintain a constant velocity. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh), where "m" is the mass of the object, "g" is the acceleration due to gravity, and "h" is the vertical distance through which the object is lowered.
As a result, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved. The work done by the external force manifests as a change in the object's gravitational potential energy, causing a decrease in potential energy as the object is lowered. This decrease in potential energy is exactly balanced by the work done by the external force, so the total mechanical energy of the system remains constant, despite the fact that kinetic energy remains constant throughout the process.
Suppose Person A is traveling on a spaceship going 50% of the speed of light. Person A measures the length of the spaceship to be 10 meters. How long would a Person B measure the spaceship if person B were on a planet as the spaceship passed by?
Answer:
L = 8.66 m
Explanation:
The length measured by the moving observer is related to the true length is given by
L = L₀ √1 - (v²/c²)
Where L₀ is the length of the spaceship as measured by person A, v is the speed of spaceship of person A and c is the speed of light c = 3.8x10ᵃ m/s
L = 10√1 - (0.5c)²/c²
L = 10√1 - (0.5*3.8x10ᵃ)²/3.8x10ᵃ ²
L = 8.66 m
Therefore, the Person B would measure the spaceship length to be 8.66 m
An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals with a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance.
What is the resistance of a 9.0 V battery that produces a 17 A current when shorted by a wire of negligible resistance?
R=____
Answer:
Resistance, [tex]R=0.529\ \Omega[/tex]
Explanation:
Given that,
Voltage of the battery, V = 9 volts
Current produced in the circuit, I = 17 A
We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :
[tex]V=IR[/tex]
[tex]R=\dfrac{V}{I}[/tex]
[tex]R=\dfrac{9\ V}{17\ A}[/tex]
[tex]R=0.529\ \Omega[/tex]
So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.
A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked together with a speed of 4 m/s. How much kinetic energy is lost in the collision?
Kinetic energy lost in collision is 10 J.
Explanation:
Given,
Mass, [tex]m_{1}[/tex] = 4 kg
Speed, [tex]v_{1}[/tex] = 5 m/s
[tex]m_{2}[/tex] = 1 kg
[tex]v_{2}[/tex] = 0
Speed after collision = 4 m/s
Kinetic energy lost, K×E = ?
During collision, momentum is conserved.
Before collision, the kinetic energy is
[tex]\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2[/tex]
By plugging in the values we get,
[tex]KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\[/tex]
K×E = 50 J
Therefore, kinetic energy before collision is 50 J
Kinetic energy after collision:
[tex]KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)[/tex]
[tex]KE = 40J + KE(lost)[/tex]
Since,
Initial Kinetic energy = Final kinetic energy
50 J = 40 J + K×E(lost)
K×E(lost) = 50 J - 40 J
K×E(lost) = 10 J
Therefore, kinetic energy lost in collision is 10 J.