Answer:
180 m
Explanation:
Case 1.
U = 40 km/h = 11.1 m/s, V = 0, s = 20 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0 = 11.1 × 11.1 - 2 × a × 20
a = 3.08 m/s^2
Case 2.
U = 220 km/h = 33.3 m/s, V = 0
a = 3.08 m/s^2
Let the stopping distance be x.
Again use third equation of motion
0 = 33.3 × 33.3 - 2 × 3.08 × x
X = 180 m
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.
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.
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.
Based on the assumption that a liquid conducting core and rapid rotation are both required for a magnetic field to operate, which terrestrial planets would you expect to have magnetic fields?
answer is earth. Earth is having everything that is required for a magnetic field to operate.
Based on the assumption that a liquid conducting core and rapid rotation are both required for a magnetic field to operate, only Earth have magnetic fields.
What are terrestrial planets?Because of their compact, rocky surfaces akin to Earth's terra firma, the planets Mercury, Venus, Earth, and Mars are referred to as terrestrial. The four planets closest to the sun are the terrestrial planets. None of the terrestrial planets have rings, although Earth does have radiation belts that have been trapped.
Only Earth possesses a sizable planetary magnetic field among the terrestrials. There is no global magnetic field on Mars or the moon of Earth, although there are localised regional magnetic fields at various locations across their surfaces.
Venus, Earth, and Mars are the only terrestrial planets with noticeable atmospheres.
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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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A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.40 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum (Let up be in the positive direction.)
Answer:
Change in momentum is 1.1275 kg-m/s
Explanation:
It is given that,
Mass of the ball, m = 274 g = 0.274 kg
It hits the floor and rebounds upwards.
The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s (-ve because the ball hits the ground)
It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)
We have to find the change in the ball's momentum. It is given by :
[tex]\Delta p=p_f-p_i[/tex]
[tex]\Delta p=m(v-u)[/tex]
[tex]\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))[/tex]
[tex]\Delta p=1.1275\ kg-m/s[/tex]
So, the change in the momentum is 1.1275 kg-m/s
The magnitude of the change in the ball's momentum when rebounding off the floor is 1.1275 kg·m/s, accounting for the change in direction during impact.
Explanation:To determine the magnitude of the change in the ball's momentum, you should first consider the initial and final momenta of the ball. Momentum is calculated as the product of mass and velocity. When the ball hits the floor, it has a downward momentum of (mass × velocity before hitting the floor). After rebounding, it has an upward momentum of (mass × velocity after rebounding). Since the problem states that up is in the positive direction, you will have to take into account the change in direction when calculating the change in momentum.
To calculate the magnitude of the change in momentum (Δp), you use the formula Δp = p_final - p_initial. Plugging in the values:
p_initial = mass × velocity before hitting = 0.275 kg × (-2.40 m/s) = -0.66 kg·m/s
p_final = mass × velocity after rebounding = 0.275 kg × 1.70 m/s = 0.4675 kg·m/s
Δp = p_final - p_initial = 0.4675 kg·m/s - (-0.66 kg·m/s) = 1.1275 kg·m/s
The negative sign for p_initial indicates that it was directed downwards. The magnitude of the change in momentum is simply the absolute value of Δp, which is 1.1275 kg·m/s.
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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.
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]
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
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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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
A child is sliding on a sled at 1.5 m/s to the right. You stop the sled by pushing on it for 0.60 s in a direction opposite to its motion. If the mass of the child and sled is 41 kg, what is the magnitude of the average force you need to apply to stop the sled? Use the concepts of impulse and momentum. Express your answer with the appropriate units.
Answer:
102.5 N
Explanation:
The impulse theorem applied to this situation states that:
[tex]F \Delta t = m \Delta v[/tex]
where
F is the average force applied on the child and the sled
[tex]\Delta t[/tex] is the time interval during which the force is applied
The term on the right represents the variation of momentum, which is the product of:
m is the mass of the child+sled
[tex]\Delta v[/tex] is the change in velocity of the child+sled
In this situation we have:
[tex]\Delta v = 0 - 1.5 m/s = -1.5 m/s[/tex]
m = 41 kg
[tex]\Delta t = 0.60 s[/tex]
So we can solve to find the average force:
[tex]F=\frac{m\Delta v}{\Delta t}=\frac{(41 kg)(-1.5 m/s)}{0.60 s}=-102.5 N[/tex]
And the negative sign means the force is applied against the direction of motion of the child. So the magnitude of the force is 102.5 N.
To stop a sled with a mass of 41 kg sliding at 1.5 m/s within 0.60 s, one needs to apply an average force of 102.5 N opposite to its direction of motion, using the concepts of impulse and momentum.
To calculate the magnitude of the average force needed to stop a sled, using the concepts of impulse and momentum. The initial velocity of the child and sled is 1.5 m/s to the right, with the mass being 41 kg, and the time over which the force is applied is 0.60 s.
Impulse equals the change in momentum, so the initial momentum (pi) is mass times velocity (41 kg x 1.5 m/s), and the final momentum (pf) is 0 (since the sled stops). Therefore, the change in momentum (Δp) is simply the initial momentum. The impulse (•Ft) equals the force times the time interval (0.60 s), equals the change in momentum. So, we can solve for force (•F) using •F = Δp / t.
The calculation is as follows: Initial momentum = 41 kg x 1.5 m/s = 61.5 kg·m/s. Since the final momentum is 0, the change in momentum (Δp) is -61.5 kg·m/s (negative, indicating a direction opposite to the initial motion). So, the magnitude of the average force needed is |Δp| / t = |(-61.5 kg·m/s) / (0.60 s)| = 102.5 N.
Therefore, the magnitude of the average force you need to apply to stop the sled is 102.5 N.
Supervisors are subject to disciplinary action for engaging in retaliation.
True
False
This statement is true.
Yes, supervisors can face disciplinary action for engaging in retaliation. They can be held accountable if they retaliate against an employee for reporting violations or participating in protected activities. Penalties can vary from warnings to termination.
Explanation:True. Supervisors, just like any other employees, are subject to disciplinary action for engaging in retaliation. For instance, if a supervisor retaliates against an employee for reporting a violation of company policies or for engaging in protected activities like organizing or supporting a labor union, they can be held accountable. Disciplinary actions can range from written warnings to termination, depending on the severity of the retaliation. Therefore, it is essential for supervisors to respect the rights of the employees and abide by all workplace regulations to maintain a safe and fair environment.
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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.
How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?
Answer:
Required time, t = 0.84 seconds
Explanation:
It is given that,
Initial speed of an object, u = 22 m/s
Final velocity of an object, v = 27 m/s
Acceleration, a = 5.93 m/s²
We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:
[tex]v=u+at[/tex]
Where
t = time
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}[/tex]
t = 0.84 seconds
Hence, the time required for an object is 0.84 seconds.
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 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 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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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.
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 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.
Learn more about Work done by friction here:https://brainly.com/question/30280752
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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.
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
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
Compared to a blue star in the same local cluster, the surface temperature of a red star is a) greater. b) the same. c) lower. d) not consistently any of these
Hello! My name is Zalgo and I am here to help you out on this concluding day. The answer would be C);lower. The reason it would be lower is because the hottest color of flames would be blue. Considering the way a start emits light is fire, this would be the most logical reason for it.
I hope that this helps! :P
"Stay Brainly and stay proud!" - Zalgo
(By the way, do you mind marking me as Brainliest? I'd greatly appreciate it! Thanks! X3)
A 10 kg plank 3 meters in length extends off the edge of a pirate ship so that only 0.5 m remains on the deck. This is held in place by a 200 kg crate of rum sitting on top of the end of the plank on the deck. A 70 kg pirate is being forced to walk the plank. How far does he get from the deck before the plank tips and he falls?
Answer:
sorry I've never took in this class before I will not be able to help you
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."
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
A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the linear charge density along the arc?
Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as
[tex]\lambda =\frac{Q}{L}[/tex]
Inserting the values
[tex]\lambda =\frac{-564.8\times 10^{-19}}{0.045}[/tex]
λ = - 1.3 x 10⁻¹⁵ C/m