Answers
Explanation:
The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by
[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]
[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]
[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]
The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by
[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]
[tex]\:\:\:\:= 26.1°[/tex]
Related Questions
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answers
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
1:
Forces and Motion:Question 2
A car is travelling east, when suddenly a more massive car travelling
north hits it with a greater force. What is likely to happen to the car
that was originally travelling east?
Answers
Explanation:
the car will be brought back
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Answers
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answers
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point
Answers
Answer:
1.37 mm²
Explanation:
From the image attached below:
Let's take a look at the two rays r and r' hitting the same mirror from two different positions.
Let x be the distance between these rays.
[tex]d_o =[/tex] distance between object as well as the mirror
[tex]d_{eye}[/tex] = distance between mirror as well as the eye
Thus, the formula for determining the distance between these rays can be expressed as:
[tex]x = 2d_o tan \theta[/tex]
where; the distance between the eye of the observer and the image is:
[tex]s = d_o + d_{eye}[/tex]
Then, the tangent of the angle θ is:
[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]
replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:
[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]
[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]
[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]
x = (0.13157 × 10) mm
x = 1.32 mm
Finally, the area A = π r²
[tex]A = \pi(\frac{x}{2})^2[/tex]
[tex]A = \pi(\frac{1.32}{2})^2[/tex]
A = 1.37 mm²
Match the following properties to the type of wave.
Answers
Answer:
hi there
Explanation:
1 - III
2- 1
3-1
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Explanation:
1 . 3
2. 1
3. 2
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5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answers
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answers
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?
Answers
Answer:
The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answers
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day
Answers
what do you mean about it
A block of mass 2 kg is launched by compressing a spring of force constant 1200 N/m. The block slides on a frictionless surface, up a 1 meter tall ramp, then it enters a region of rough surface. It comes to a stop after traveling 3 meters over the rough surface. The coefficient of kinetic friction between the block and the rough surface is 0.40.
Required:
a. How many forces end up doing work on the block from release to stop?
b. What is the total non-conservative work done on the block?
c. What is the change in the spring potential energy of the block?
Answers
Answer:
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Explanation:
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answers
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
calculate the length of wire.
Answers
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answers
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answers
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
A capacitor consists of two parallel conducting plates, each of area 0.4 m2 and separated by a distance of 2.0 cm. Assume there is air between the plates. While connected to a battery the electric field within the plates is 500 N/C. The potential difference between the plates is: ________
a) 5.0 V
b) 10 V
c) 30 V
d) 20 V
Answers
Answer:
check photo
Explanation:
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
Answers
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answers
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
What type of wave is a microwave?
O heat
O longitudinal
sound
transverse
Answers
Answer:
Microwave is a types of a electromagnetic radiation
Answer:
Transvers
Explanation:
Because microwave is electromagnetic waves and all electromagnetic waves are transvers.
Why is it advised not to hold the thermometer by its bulb while reading it?
Answers
Q)what are convex mirrors?
Answers
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement
Answers
Answer:
|x| = √53
Explanation:
We are told that the vector starts at the point (0.0) and ends at (2,-7) .
Thus, magnitude of displacement is;
|x| = √(((-7) - 0)² + (2 - 0)²)
|x| = √(49 + 4)
|x| = √53
1. Why do only some people get addicted to
drugs?
Answers
Answer:
When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answers
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
what is the distance time how can we find the speed of an object from its distance time graph
Answers
Answer:
speed is the gradient of the graph
Answer:
Speed is the slope of a distance time graph.
Explanation:
Speed= d/t
Slope is equal to rise/run
If the rise of the graph is the distance and the run is the time, calculating slope is the equivalent of calculating average speed.
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answers
Answer:
b = frequency
The cation that is reabsorbed from the urine in response to aldosterone
Answers
Answer:
If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.
if you jog at a speed of 1.5m/s for 20 seconds how far di you travel
Answers
Answer: 30m
Explanation:
Given:
Speed: 1.5m/s
Time: 20 seconds
Distance = speed × time
Distance = 1.5 × 20
= 30m
Therefore you will travel 30m
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Explain how blood circulation takes place in humans?
Answers
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
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