Answer:
No
Explanation:
When the person slides down, the change in gravitational potential energy is converted into kinetic energy, according to
[tex]\Delta U = \Delta K\\mg\Delta h = \frac{1}{2}mv^2[/tex]
where
m is the mass of the person
g is the acceleration of gravity
v is the final speed
[tex]\Delta h[/tex] is the change in heigth of the person
Here we have assumed that the initial speed is zero.
Re-arranging the equation,
[tex]v = \sqrt{2g \Delta h}[/tex]
and we see that this quantity does not depend on the mass of the person, so every person will have the same speed at the bottom of the slide, equal to:
[tex]v=\sqrt{2(9.8 m/s^2)(3.2 m-1.2 m)}=6.3 m/s[/tex]
Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N
(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:
[tex]I=\Delta p = m\Delta v[/tex]
where
m = 62.0 kg is the mass of the passenger
[tex]\Delta v[/tex] is the change in velocity of the car (and the passenger), which is
[tex]\Delta v = 5.30 m/s - 0 = 5.30 m/s[/tex]
So, the linear impulse experienced by the passenger is
[tex]I=(62.0 kg)(5.30 m/s)=328.6 kg m/s[/tex]
(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:
[tex]I=F \Delta t[/tex]
where in this case
[tex]I=328.6 kg m/s[/tex] is the linear impulse
[tex]\Delta t = 0.812 s[/tex] is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:
[tex]F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N[/tex]
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.15 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 35.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target? m/s
Answer:
6.83 m/s
Explanation:
Momentum is conserved.
Initial momentum = final momentum
(300 g) (-2.15 m/s) + (22.5 g) (35.5 m/s) = (22.5 g) v
v = 6.83 m/s
You have a perfectly circular apple pie with a radius of 15cm. You cut the pie into 12 approximately equal slices. What is the arc length (linear length) and angular displacement of each slice's crust?
Answer:
7.85 cm
30 degree
Explanation:
As we know that that the angular displacement in one turn is 360 degree.
As there are 12 parts which are equally divided so the angle turn by each part is
= 360 / 12 = 30 degree
Thus, the angular displacement of each slice is 30 degree.
Radius of pie = 15 cm
Circumference of pie = 2 x 3.14 x 15 = 94.2 cm
Arc length for complete 360 degree = 94.2 cm
Arc length for 30 degree turn = 94.2 x 30 / 360 = 7.85 cm
An object whose mass is 100 lb falls freely under the influence of gravity from an initial elevation of 600 ft above the surface of Earth. The initial velocity is downward with a magnitude of 50 ft/s. The effect of air resistance is negligible. Determine the velocity, in ft/s, of the object just before it strikes Earth. Assume g = 31.5 ft/s
Answer with Explanations:
Given:
Mass of object, m = 100 lb
height fallen, h = 600 ft
initial velocity, u = 50 ft/s
acceleration due to gravity, g = 31.5 ft/s^2
Find final velocity when it touches ground.
Solution:
Use standard kinematics equation, in the absence of air resistance and variation of g with height,
v^2 - u^2 = 2aS
where
v = final velocity
u = initial velocity
a = acceleration due to gravity
S = distance travelled
Substitute values
v^2 = u^2 + 2aS
= 50^2 + 2*31.5*600
= 40300 ft^2/s^2
Final velocity,
v = sqrt(40300) ft/s
= 200.75 ft/s
= 201 ft/s to the nearest foot.
A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the acceleration of the block. 1.23 Correct: Your answer is correct. m/s2 (b) Find the coefficient of kinetic friction between the block and the incline. .23 Incorrect: Your answer is incorrect. (c) Find the frictional force acting on the block. N (d) Find the speed of the block after it has slid 2.00 m.
(a) [tex]1.23 m/s^2[/tex]
Let's analyze the motion along the direction of the incline. We have:
- distance covered: d = 2.00 m
- time taken: t = 1.80 s
- initial velocity: u = 0
- acceleration: a
We can use the following SUVAT equation:
[tex]d = ut + \frac{1}{2}at^2[/tex]
Since u=0 (the block starts from rest), it becomes
[tex]d=\frac{1}{2}at^2[/tex]
So by solving the equation for a, we find the acceleration:
[tex]a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2[/tex]
(b) 0.50
There are two forces acting on the block along the direction of the incline:
- The component of the weight parallel to the surface of the incline:
[tex]W_p = mg sin \theta[/tex]
where
m = 3.00 kg is the mass of the block
g = 9.8 m/s^2 is the acceleration due to gravity
[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline
This force is directed down along the slope
- The frictional force, given by
[tex]F_f = - \mu mg cos \theta[/tex]
where
[tex]\mu[/tex] is the coefficient of kinetic friction
According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:
[tex]W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma[/tex]
Solving for [tex]\mu[/tex], we find
[tex]\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50[/tex]
(c) 12.3 N
The frictional force acting on the block is given by
[tex]F_f = \mu mg cos \theta[/tex]
where
[tex]\mu = 0.50[/tex] is the coefficient of kinetic friction
m = 3.00 kg is the mass of the block
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline
Substituting, we find
[tex]F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N[/tex]
(d) 6.26 m/s
The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation
[tex]v^2 - u^2 = 2ad[/tex]
where
v is the final speed of the block
u = 0 is the initial speed
a = 1.23 m/s^2 is the acceleration
d = 2.00 m is the distance covered
Solving the equation for v, we find the speed of the block after 2.00 m:
[tex]v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s[/tex]
The acceleration of the block is 0.62 m/s². The coefficient of kinetic friction between the block and the incline is 0.048.
Explanation:First, let's find the acceleration of the block. To do this, we can use the second law of motion, which states that acceleration equals the net force divided by the mass. But before we can find the net force, we need to identify the individual forces at play. The force of gravity acting on the object is 3.00 kg * 9.8 m/s² = 29.4 N. This force acts vertically downward, but since the block is on an incline, we have to resolve this force into two components: one parallel to the incline and one perpendicular to the incline. The parallel component, which is the force that actually moves the block, equals Fg * sin(33.0°) = 29.4 N * sin(33.0°) = 16.14 N. Since the block starts at rest and then speeds up, it must be accelerating. We can calculate that acceleration with the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. The problem tells us the block slides 2.00 m in 1.80 s, so we can calculate Δv using the formula Δv = Δd / Δt. Substituting the given values gives us Δv = 2.00 m / 1.80 s = 1.11 m/s. Therefore, a = Δv / Δt = 1.11 m/s / 1.80 s = 0.62 m/s².
Now let's find the coefficient of kinetic friction between the block and the incline. We know that the force of friction equals the force of gravity component perpendicular to the incline times the coefficient of kinetic friction (f = μk * Fg * cos(33.0°)), and this is equal to the force of gravity component parallel to the incline minus the force that results from the block's acceleration (f = Fg * sin(33.0°) - m*a). In other words, μk = (Fg * sin(33.0°) - m*a) / (Fg * cos(33.0°)). Substituting the given values gives us μk = (29.4 N * sin(33.0°) - 3.00 kg * 0.62 m/s²) / (29.4 N * cos(33.0°)) = 0.048.
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Given the position vector of the particle
r(t)=(t+1)i+(t^2−1)j+2t k, find the particle's velocity and acceleration vectors at t=1
With position vector
[tex]\vec r(t)=(t+1)\,\vec\imath+(t^2-1)\,\vec\jmath+2t\,vec k[/tex]
the particle then has velocity
[tex]\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\vec\imath+2t\,\vec\jmath+2\,vec k[/tex]
and acceleration
[tex]\vec a(t)=\dfrac{\mathrm d\vec v(t)}{\mathrm dt}=\dfrac{\mathrm d^2\vec r(t)}{\mathrm dt^2}=2\,\vec\jmath[/tex]
Then [tex]t=1[/tex], then particle's velocity and acceleration are, respectively,
[tex]\vec v=\vec\imath+2\vec\jmath+2\,\vec k[/tex]
and
[tex]\vec a=2\,\vec\jmath[/tex]
The maximum force measured on a 3.72 m wire is 0.731 N when the magnetic field is 0.093 T. What current flows through the wire? O 2.36 A O 225A O 2.11 A O 218
Answer:
Current, I = 2.11 A
Explanation:
It is given that,
Length of the wire, L = 3.72 m
Maximum force on the wire, F = 0.731 N
Magnetic field, B = 0.093 T
We have to find the current flowing the wire. The force acting on the wire is given by :
[tex]F=lLB\ sin\theta[/tex]
When [tex]\theta=90[/tex], F = maximum
So, [tex]F=ILB[/tex]
[tex]I=\dfrac{F}{LB}[/tex]
[tex]I=\dfrac{0.731\ N}{3.72\ m\times 0.093\ T}[/tex]
I = 2.11 A
So, the current flowing through the wire is 2.11 A. Hence, the correct option is (c).
A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80 m/s. What is the momentum of the 202 kg car afterwards? (Unit-kg*m/s)
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s[/tex]
p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
what is the momentum of a 48.2N bowling ball with a velocity of 7.13m/s?
Answer:
Momentum, p = 34.937 kg-m/s
Explanation:
It is given that,
Force acting on the bowling ball, F = 48.2 N
Velocity of bowling ball, v = 7.13 m/s
We have to find the momentum of the ball. Momentum is given by :
p = mv........(1)
Firstly, calculating the mass of bowling ball using second law of motion. The force acting on the ball is gravitational force and it is given by :
F = m g (a = g)
[tex]m=\dfrac{F}{g}[/tex]
[tex]m=\dfrac{48.2\ N}{9.8\ m/s^2}[/tex]
m = 4.9 kg
Now putting the value of m in equation (1) as :
[tex]p=4.9\ kg\times 7.13\ m/s[/tex]
p = 34.937 kg-m/s
Hence, this is the required solution.
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s2 and there is no atmosphere. How long does it take for the instrument to return to where it was thrown?
Explanation:
The situation described here is parabolic movement. However, as we are told the instrument is thrown upward from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
[tex]y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}[/tex] (1)
Where:
[tex]y[/tex] is the instrument's final position
[tex]y_{o}=0[/tex] is the instrument's initial position
[tex]V_{o}=15m/s[/tex] is the instrument's initial velocity
[tex]t[/tex] is the time the parabolic movement lasts
[tex]g=2.5\frac{m}{s^{2}}[/tex] is the acceleration due to gravity at the surface of planet X.
As we know [tex]y_{o}=0[/tex] and [tex]y=0[/tex] when the object hits the ground, equation (1) is rewritten as:
[tex]0=V_{o}.t-\frac{1}{2}g.t^{2}[/tex] (2)
Finding [tex]t[/tex]:
[tex]0=t(V_{o}-\frac{1}{2}g.t^{2})[/tex] (3)
[tex]t=\frac{2V_{o}}{g}[/tex] (4)
[tex]t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}[/tex] (5)
Finally:
[tex]t=12s[/tex]
Final answer:
Using the kinematic equation for free fall, the time it takes for the instrument to reach the point of zero velocity on planet X is 6 seconds. Since the descent takes an equal amount of time as the ascent, the total round trip time is 12 seconds.
Explanation:
To determine the time it takes for the instrument to return to its original position, we can use the kinematic equation for free fall motion under uniform acceleration, which is given by:
v = u + at
Where:
v is the final velocity (0 m/s at the highest point)u is the initial velocity (15 m/s)a is the acceleration due to gravity (-2.5 m/s^2; negative because it's opposite the direction of initial velocity)t is the timeRearranging the equation to solve for t:
t = (v - u) / a
The time it takes to reach the highest point is:
t = (0 m/s - 15 m/s) / (-2.5 m/s^2) = 6 seconds
To find the total time for the round trip, we need to double this time because the descent will take the same amount of time as the ascent:
Total time = ascent time + descent time = 6 s + 6 s = 12 seconds.
what is the critical angle for light going from crown glass to air?
Answer:
The critical angle for a diamond in air is 24 degrees, while the critical angle for glass is 41 degrees.
Explanation:
Rays exiting the material at an angle less than the critical angle will be refracted, and rays incident on the interface at greater than the critical angle will be totally reflected back inside the material.
A satellite with mass 500 kg is placed in a circular orbit about Earth (Mass= 5.98 x 10^24 kg), radius = (6.4 x 10^6), a distance of 1500 km above the surface. (a) what is the force gravity acting on satellite? (b) what is the satellite’s acceleration? (c) what is the satellite’s orbital speed?
Explanation:
a) F = GmM / r²
F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²
F = 3200 N
b) F = ma
3200 = 500a
a = 6.4 m/s²
c) a = v² / r
640 = v² / (6.4×10⁶ + 1.5×10⁶)
v = 7100 m/s
In a game of football outdoors on a cold day, a player will begin to feel exhausted after using approximately 8.0 × 105 J of internal energy. (a) One player, dressed too lightly for the weather, has to leave the game after losing 6.8 × 105 J of heat. How much work has he done? (b) Another player, wearing clothes that offer better protection against heat loss, is able to remain in the game long enough to do 2.4 × 105 J of work. What is the magnitude of the heat that he has lost?
Answer:
Part a)
[tex]W = 1.2 \times 10^5 J[/tex]
Part b)
[tex]Q = 5.6 \times 10^5 J[/tex]
Explanation:
It given that player will feel exhausted when he is using his internal energy of [tex]8.0 \times 10^5 J[/tex]
PART a)
it is given that heat loss by the player is given as
[tex]Q = 6.8 \times 10^5 J[/tex]
now by first law of thermodynamics we have
[tex]\Delta U = Q + W[/tex]
now we have
[tex]8.0 \times 10^5 = 6.8 \times 10^5 + W[/tex]
[tex]W = 1.2 \times 10^5 J[/tex]
PART b)
It is given that another player did the work as
[tex]W = 2.4 \times 10^5 J[/tex]
now we have first law of thermodynamics
[tex]\Delta U = Q + W[/tex]
now we have
[tex]8.0 \times 10^5 = 2.4 \times 10^5 + Q[/tex]
[tex]Q = 5.6 \times 10^5 J[/tex]
Using the first law of thermodynamics, we find that the first player has done -1.2 x 10^5 J of work and the second player has lost 10.4 x 10^5 J of heat.
Explanation:The questions posed are about applying the concept of the first law of thermodynamics in determining the amount of work done by a football player and the heat lost in the process. This law is also known as the law of energy conservation, and it can be written as ΔU = Q - W, where ΔU is the change in the system's internal energy, Q is the heat added to the system, and W is the work done by the system.
(a) The player who was dressed too lightly had to leave the game after losing 6.8 × 105 J of heat. Using the first law of thermodynamics, we can calculate the work done by the player. If the change in the internal energy before the player gets exhausted is 8.0 × 105 J (the energy used) and the heat lost is 6.8 × 105 J, then the work done (W) can be calculated as follows:
W = Q - ΔU = 6.8 × 105 J - 8.0 × 105 J = -1.2 × 105 J
(b) Another player was able to do 2.4 × 105 J of work before getting exhausted. The magnitude of the heat that he lost can then be calculated as follows:
Q = ΔU + W = 8.0 × 105 J + 2.4 × 105 J = 10.4 × 105 J
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Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of time are x ( t ) = 8 t and y ( t ) = − 16 t 2 + 100 and . How long does it take her to reach the ground? How far away from the cliff is she when she lands?
It takes Lukalu 2.5 seconds to reach the ground, and she lands 20 meters away from the cliff.
To determine how long it takes Lukalu to reach the ground, we need to find the value of [tex]\( t \)[/tex] when [tex]\( y(t) = 0 \)[/tex], since[tex]\( y(t) \)[/tex] represents her vertical position. The parametric equation for \( y(t) \) is given by[tex]\( y(t) = -16t^2 + 100 \).[/tex] Setting [tex]\( y(t) \)[/tex] equal to zero gives us the equation:
[tex]\[ -16t^2 + 100 = 0 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ 16t^2 = 100 \] \[ t^2 = \frac{100}{16} \] \[ t^2 = 6.25 \] \[ t = \sqrt{6.25} \] \[ t = 2.5 \][/tex]
So, it takes Lukalu 2.5 seconds to reach the ground.
Next, to find out how far away from the cliff she is when she lands, we need to evaluate [tex]\( x(t) \) at \( t = 2.5 \)[/tex] seconds. The parametric equation for [tex]\( x(t) \)[/tex] is given by [tex]\( x(t) = 8t \)[/tex]. Plugging in the value of [tex]\( t \),[/tex] we get:
[tex]\[ x(2.5) = 8 \times 2.5 \][/tex]
[tex]\[ x(2.5) = 20 \][/tex]
Therefore, Lukalu is 20 meters away from the cliff when she lands."
A diver in midair has an angular velocity of 6.0 rad/s and a moment of inertia of 1.2 kg·m2. He then pulls is arms and legs into a tuck position and his angular velocity increases to 12 rad/s. The net external torque acting on the diver is zero. What is his moment of inertia in the tuck position?
Answer:
0.6 kg m²
Explanation:
Angular momentum is conserved.
Iω = Iω
(1.2 kg m²) (6.0 rad/s) = I (12 rad/s)
I = 0.6 kg m²
The diver's moment of inertia in the tuck position is found to be 0.6 kg·m^2 using the conservation of angular momentum, given that no external torque acts on him.
Explanation:The question relates to the concept of conservation of angular momentum, which is a principle in physics stating that if no external torque acts on a system, the total angular momentum of the system remains constant. In this problem, a diver's angular velocity increases as he changes from a relaxed position to a tucked position, indicating that his moment of inertia must decrease to conserve angular momentum because external torque is zero.
To solve for the diver's moment of inertia in the tucked position, we use the formula for conservation of angular momentum:
L_initial = L_finalI_initial * ω_initial = I_final * ω_final(1.2 kg·m2) * (6.0 rad/s) = I_final * (12 rad/s)I_final = (1.2 kg·m2 * 6.0 rad/s) / 12 rad/sI_final = 0.6 kg·m2Therefore, the moment of inertia of the diver in the tuck position is 0.6 kg·m2.
A 6.0-kilogram block slides along a horizontal surface. If μk = 0.20 for the block and surface, at what rate is the friction force doing work on the block at an instant when its speed is 4.0 m/s?
Answer:
Power = 47.0 Watt
Explanation:
As we know that friction force is given by
[tex]F_f = \mu mg[/tex]
now we have
[tex]\mu = 0.20[/tex]
m = 6.0 kg
now we have
[tex]F_f = 0.20(6.0)(9.80) = 11.76 N[/tex]
now since we need to find the rate of work done by friction force
so we can say rate of work done is power due to friction force
so it is given as
[tex]P = F_f (v)[/tex]
[tex]P = 11.76 (4.0)[/tex]
[tex]P = 47.0 Watt[/tex]
The rate at which the frictional force is doing work on the block when it's moving at a speed of 4.0 m/s is 47.04 Watts.
Explanation:The rate at which the frictional force is doing work on the 6.0-kilogram block sliding along a horizontal surface can be obtained by recognizing that work done per unit time is equal to power. The frictional force (F) acting on the block is given by F = μkN, where μk is the coefficient of kinetic friction and N is the normal force. In this case, since the surface is horizontal, N is equal to the weight of the block, which is mass (m) times gravity (g).
Therefore, F = μkmg = 0.20 * 6.0 kg * 9.8 m/s² = 11.76 N.
The power (P) done by the force of friction is given by P = Fv, where v is the velocity. So, P = 11.76 N * 4.0 m/s = 47.04 Watts.
This calculates to be the rate at which the frictional force is doing work on the block when it is moving at a speed of 4.0 m/s.
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If an object with an initial temperature of 300 K increases its temperature by 1°C every minute, by how many degrees Fahrenheit will its temperature have increased in 10 minutes? (A) 6°F (B) 10°F (C) 18°F (D) 30°F
A temperature increase of 1°C every minute translates to a temperature increase of 1.8°F every minute. Over 10 minutes, the temperature would therefore increase by 18°F.
Explanation:The question is asking about temperature increase in an object. If an object with an initial temperature of 300 K increases its temperature by 1°C every minute, we first need to understand the connection between degrees Celsius and Fahrenheit. Namely, a difference of 1 degree Celsius is equivalent to a difference of 1.8 degrees Fahrenheit.
So, if the temperature increases by 1°C every minute, it would increase by 1.8°F every minute. If we look at a span of 10 minutes, we use simple multiplication to find the total increase. The temperature would increase by 1.8°F x 10 = 18°F over the course of 10 minutes.
Thus, the answer is (C) 18°F.
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Which of the following is a conservative force? 1.force due to friction 2.force due to gravity 3. both (1) and (2) 4. neither (1) nor (2)
Answer:
Option 2 is the correct answer.
Explanation:
I f the work done by a force does not depend upon the path of mass then the force is called conservative force.
Work done by frictional force depends upon path followed by mass, so frictional force is a non conservative force. But work done by gravitational force does not depend upon path followed by mass, so gravitational force is a conservative force.
Option 2 is the correct answer.
if a load of 1000kg can just be dragged up an incline at 10 degrees to the horizontal by a force of 5KN applied in the most effective direction,what is the value offthe coefficient offriction?
Answer:
The coefficient of friction is 0.34
Explanation:
It is given that,
Mass of the load, m = 1000 kg
It is dragged up an incline at 10 degrees to the horizontal by a force of 5 KN applied in the most effective direction, F = 5 × 10³ N
We need to find the coefficient of friction between the surface and the load. From the attached figure, the load is dragged up with a force of F. A frictional force f will also act in this scenario.
So, [tex]F=f+mg\ sin\theta[/tex]
Since, [tex]f=\mu N[/tex]
or [tex]f=\mu mg\ cos\theta[/tex]
[tex]F=\mu mg\ cos\theta+mg\ sin\theta[/tex]
[tex]F-mg\ sin\theta=\mu mg\ cos\theta[/tex]
[tex]5\times 10^3\ N-1000\ kg\times 9.8\ m/s^2\ sin(10)=\mu mg\ cos\theta[/tex]
[tex]\mu=\dfrac{3298.24}{1000\ kg\times 9.8\ m/s^2\times cos(10)}[/tex]
[tex]\mu=0.34[/tex]
So, the coefficient of friction is 0.34. Hence, this is the required solution.
What was the average force exerted on a 44 kg ojbect if the inital velocity was 7 m/s, the final velocity was 2 m/s and t was exerted for 4.5 s. Leave no spaces between units and the answer. No decimals.
Answer:
Force exerted = 48.89 N
Explanation:
Force = Mass x Acceleration
Mass = 44 kg
Acceleration is rate of change of velocity.
Acceleration, [tex]a=\frac{2-7}{4.5}=-1.11m/s^2[/tex]
Force = Mass x Acceleration = 44 x -1.11 = -48.89 N
Force exerted = 48.89 N
Bank robbers have pushed a 1000 kg safe to a second story floor to ceiling window. They plan to break the window, then lower the safe 3.0 meteres to their truck. Not being too clever, they stack up 500 kg of furnature, tie a rope between the safe and the furnature, and place the rope over the pully. Then they push the safe out of the window. what is the saves speed when it hits the truck? The coefficient of kinetic friction between the furniture and the floor is 0.50.
Explanation with answer:
First, in problems like this, it is always clear to draw a diagram to make sure you understand the problem. If it is not possible to draw the diagram correctly, perhaps something is misunderstood or missing from the question.
Here, see the attached image.
Note that the rope has a tension of T that pulls both the furniture and the safe.
To find the final speed (when the safe hits the truck), we need first to find the acceleration.
The system's total mass, M = 1000+500 kg = 1500 kg
Forces acting on the system
= gravity acting on the safe less friction acting on the furniture.
= m1*g - mu*m2g
= 1000*9.81 - 0.5*500*9.81
= 7357.5 N
Acceleration, a = F/m = 7357.5 / 1500 = 4.905 m/s^2
Initial speed = 0 m/s
distance travelled, S = 3m
Let final speed = v
Kinematics equation gives
v^2-u^2 = 2aS
v^2 = 2*4.905*3 - 0^2 = 29.43 m^2/s^2
final speed, v = sqrt(29.43) = 5.4 m/s (to two significant figures.
The safe's speed when it hits the truck can be determined by considering the conversion of its initial potential energy to kinetic energy, and subtracting the work done to overcome friction.
Explanation:To determine the safe's speed when it hits the truck, we apply principles of conservation of energy and account for the work done by friction forces. Initially, the gravitational potential energy of the safe is given by mgh = 1000 kg * 9.8 m/s² * 3.0 m = 29400 J. As the safe drops, its potential energy is converted to kinetic energy (0.5*mv²) while energy is also consumed to overcome the frictional force on the furniture.
The friction force is μK * N = 0.5 * 500 kg * 9.8 m/s² = 2450 N. The work done by this force over 3.0 m is 2450 N * 3.0 m = 7350 J.
As energy is conserved, the kinetic energy of the safe when it hits the truck will be the initial potential energy minus the work done on friction. So, 0.5 * 1000 kg * v² = 29400 J - 7350 J. Solving this equation will give you the speed v of the safe when it hits the truck.
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(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs? [Hint: Assume roughly that brightness is proportional to power consumed.]
Answer:
Explanation:
You are looking for the resistance to start with
W = E * E/R
75 = 240 * 240 / R
75 * R = 240 * 240
R = 240 * 240 / 75
R = 57600 / 75
R = 768
Now let's see what happens when you try putting this into 110
W = E^2 / R
W = 120^2 / 768
W = 18.75
So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of
18.75 / 75 = 0.25
This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25
Final answer:
The brightness of the 75-W 240 V bulb relative to the 75-W 120 V bulb is 50%.
Explanation:
When comparing the brightness of a 75-W lightbulb operating at 240 V in Europe to a 75-W lightbulb operating at 120 V in the United States, we can use the fact that brightness is proportional to power consumed. Since power is equal to voltage multiplied by current, we can calculate the current for each bulb using the formula P = IV. For the 75-W 240 V bulb, the current is 0.3125 A, and for the 75-W 120 V bulb, the current is 0.625 A. The brightness of the European bulb relative to the US bulb can be calculated by dividing the current of the European bulb by the current of the US bulb: 0.3125 A / 0.625 A = 0.5, or 50%.
An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector inside a TV tube. The electron strikes the screen after traveling 11 cm horizontally and 34 cm vertically upward with no horizontal acceleration. What is the constant vertical acceleration provided by the deflector? (The effects of gravity can be ignored.)
Answer:
a = 5.05 x 10¹⁴ m/s²
Explanation:
Consider the motion along the horizontal direction
[tex]v_{x}[/tex] = velocity along the horizontal direction = 3.0 x 10⁶ m/s
t = time of travel
X = horizontal distance traveled = 11 cm = 0.11 m
Time of travel can be given as
[tex]t = \frac{X}{v_{x}}[/tex]
inserting the values
t = 0.11/(3.0 x 10⁶)
t = 3.67 x 10⁻⁸ sec
Consider the motion along the vertical direction
Y = vertical distance traveled = 34 cm = 0.34 m
a = acceleration = ?
t = time of travel = 3.67 x 10⁻⁸ sec
[tex]v_{y}[/tex] = initial velocity along the vertical direction = 0 m/s
Using the kinematics equation
Y = [tex]v_{y}[/tex] t + (0.5) a t²
0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²
a = 5.05 x 10¹⁴ m/s²
The vertical acceleration provided by the deflector is 5.05 x 10¹⁴ m/s².
What is acceleration?
Acceleration can be defined as the change in speed or direction of the object or particle.
First, calculate the time for horizontal motion,
[tex]t = \dfrac d v_x[/tex]
Where,
[tex]d[/tex] - horizontal distance = 11 cm = 0.11 m
[tex]v_x[/tex] - horijontal velocity = 3.0 x 10⁶ m/s
So,
t = 3.67 x 10⁻⁸ sec
Now calculate for vertical acceleration,
[tex]Y = v^o\times t + (0.5) a t^2[/tex]
Where,
Y - verical distance = 34 cm = 0.34 m
v^o - initial vertical velocity = 0 m/s.
a - acceleration = ?
Put the values in the formula,
0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²
a = 5.05 x 10¹⁴ m/s²
Therefore, the vertical acceleration provided by the deflector is 5.05 x 10¹⁴ m/s².
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A twin-sized air mattress used for camping has dimensions of 75 cm by 225 cm by 17.5 cm when blown up. The weight of the mattress is 2 kg How heavy a person, in newtons, could the air mattress support if it is floating in freshwater, with density1000 kg/m^3?
Answer:
2870 N
Explanation:
There are three forces on the mattress. Weight of the mattress, weight of the person, and buoyancy.
∑F = ma
B - mg - Mg = 0
Buoyancy is equal to the weight of the displaced fluid.
ρVg - mg - Mg = 0
ρV - m = M
Plugging in values:
M = (1000 kg/m³) (0.75 m × 2.25 m × 0.175 m) - 2 kg
M = 293 kg
The person's weight is therefore:
Mg = 293 kg × 9.8 m/s²
Mg = 2870 N
To calculate the maximum weight (in newtons) a twin-sized air mattress can support when floating in freshwater, the buoyant force is determined by the amount of water the mattress displaces, multiplied by the density of the water. Converting displaced water weight to newtons and subtracting the weight of the mattress provides the net buoyant force, which is the maximum supportable weight.
Calculating the Buoyant Force and Supportable Weight by an Air Mattress in Freshwater
To determine how heavy a person a twin-sized air mattress can hold when placed in freshwater, we use the principle of buoyancy. Buoyancy describes the upward force exerted by a fluid that opposes the weight of an immersed object. In this case, the air mattress is the immersed object in freshwater.
The buoyant force can be found using Archimedes' principle, which states that the buoyant force is equal to the weight of the water displaced by the object. The displacement volume of the mattress can be calculated by its dimensions: 75 cm by 225 cm by 17.5 cm. However, for buoyancy calculations, we use metric units, so the dimensions should be converted to meters. The volume is thus 0.75 m x 2.25 m x 0.175 m = 0.2953125 cubic meters. The weight of the water displaced can then be calculated by multiplying this volume by the density of freshwater, which is 1000 kg/[tex]k^{3}[/tex], resulting in a displacement weight of 295.3125 kg.
To find the maximum weight the mattress can support, we convert the displacement weight to newtons (knowing that 1 kg = 9.81 N) which gives us approximately 2894.17 N. Since the mattress itself weighs 2 kg (19.62 N), the total buoyant force it can exert is the sum of the weight it can displace (2894.17 N) and its own weight. Therefore, subtracting the weight of the mattress in newtons from the total buoyant force gives us the net buoyant force, which is the maximum weight of a person it can support in newtons.
When a 1.50V battery is shorted by a copper wire whose resistance can be ignored, the current through the copper wire is 2.75 A. What is the internal resistance of the battery?
Answer:
Internal resistance = 0.545 ohm
Explanation:
As per ohm's law we know that
[tex]V = iR[/tex]
here we know that
i = electric current = 2.75 A
V = potential difference = 1.50 Volts
now from above equation we have
[tex]1.50 = 2.75 ( R)[/tex]
now we have
[tex]R = \frac{1.50}{2.75}[/tex]
[tex]R = 0.545 ohm[/tex]
Final answer:
To find the internal resistance of a 1.50V battery shorted by a copper wire with a current of 2.75 A, you use Ohm's Law. The calculation involves dividing the total voltage by the current, yielding an internal resistance of 0.545 ohms.
Explanation:
When a 1.50V battery is shorted by a copper wire whose resistance can be ignored, the current through the copper wire is 2.75 A, you are asked to find the internal resistance of the battery. This problem can be solved using Ohm's Law, which states that the voltage (V) across a resistor is the product of the current (I) passing through it and its resistance (R), denoted as V = IR. However, when considering a battery, we must account for its internal resistance (r).
In this case, the internal resistance of the battery causes a voltage drop inside the battery, which is also governed by Ohm's Law (V = Ir).
Given the total voltage (emf of the battery, 1.50 V) and the current (2.75 A), we can rearrange the formula to solve for r: r = V/I.
Substituting the given values gives us: r = 1.50V / 2.75A = 0.545 Ω.
Therefore, the internal resistance of the battery is approximately 0.545 ohms.
Calculate the focal length of the mirror formed by the convex side of a shiny spoon that has a 1.54 cm radius of curvature. (b) What is its power in diopters?
Explanation:
It is given that,
Radius of curvature of the mirror, R = 1.54 cm
(a) We have to find the focal length of the mirror. The relationship between the focal length and the radius of curvature is given by :
[tex]R=2f[/tex]
f = focal length of the mirror
[tex]f=\dfrac{1.54\ cm}{2}[/tex]
f = 0.77 cm
(b) The power of mirror is given by the reciprocal of focal length i.e.
Power, [tex]P=\dfrac{1}{f}[/tex]
P = 1.29 diopters
Hence, this is the required solution.
Consider two isolated, charged conducting spheres: a large sphere and a second smaller sphere with a radius 6 times smaller than that of the large sphere, but with 3 times as much charge.
(a) Calculate the ratio of the electric potential at the surface of the large sphere to that of the small sphere.
Let the bigger sphere be sphere 1 and the let the smaller sphere be sphere 2. Rest of the answer is in the picture.
An ideal heat pump is being considered for use in heating an environment with a temperature of 22.4°C. What is the cold reservoir temperature (in degrees C) if the pump is to have a coefficient of performance of 11.7?
Answer:
- 0.86 C
Explanation:
Let the temperature of cold reservoir is T2.
T1 = 22.4 C = 295.4 K, B = 11.7
By the formula of coefficient of performance of heat pump
B = T2 / (T1 - T2)
11.7 = T2 / (295.4 - T2)
11.7 × 295.4 - 11.7 T2 = T2
T2 = 272.14 K
T2 = - 0.86 C
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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 rev/min.
Answer:
(a) 152.85 Nm
(b) 1528.5 Nm
Explanation:
According to the formula of power
P = τ ω
ω = 2 π f
(a) f = 2500 rpm = 2500 / 60 = 41.67 rps
So, 40 x 1000 = τ x 2 x 3.14 x 41.67
τ = 152.85 Nm
(b) f = 250 rpm = 250 / 60 = 4.167 rps
So, 40 x 1000 = τ x 2 x 3.14 x 4.167
τ = 1528.5 Nm