Answer:
181.87 N
Explanation:
Given that a mountain climber lies flat on the surface of a glacier (assumed frictionless), which is inclined at 30 degrees to the horizontal. he remains stationary because he is holding on to a rope that is attached to a spoke in the ice further up the slope.
Since the he slope is parallel to the surface of the ice.
if the force of the ice on the climber is 210N, the tension T in the rope will be
T = 210 cos 30
T = 181.86 N
Therefore, the tension in the rope is 181.86 N
3. A block of mass m is suspended by strings as shown in the figure. The tension in the horizontal string
is 36 N. The angle 0 is 60°. Find the mass of the block and the tension forces in string A and string
B
I am having trouble trying to answer this problem. Any help?
Answer:
mass of the block is 7.2Kg
Tension on string A = 72N
Tension on the string with an angle is = 108N
Explanation:
The Horizontal string will have the cos component of the force...
Hence, mg cos60 = 36
by keeping acceleration due to gravity = 10 m/s^2, we get,
m = 7.2Kg
The tension in String A = mg = 7.2*10 = 72N
since the whole string holding the block is pulled by another string with 36 N
the resultant force will be the tension of the string with an angle.
Hence, 72 +36 = 108 N
Please let me know whether I got this right or not...
Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N
B. 3N
C. 8N
D. 120N
Answer:
F = 0.012 N
Explanation:
Given that,
Mass of the object, m = 6 g
Acceleration, a = 2 m/s²
1 kg = 1000 grams
6 g = 0.006 kg
Force, F = ma
So,
[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]
So, the force is 0.012 N.
Which phenomenon related to light allows scientists to measure the distance between a star and Earth? A. Dark patches appear in the electromagnetic spectrum emitted by stars. B. Light waves spread out as they move farther from their source. C. Stars have different colors based on their temperature. D. Stars have unique elements that make up their atmospheres.
Answer:
The Answer Is B not C
Explanation:
Answer:b
Explanation:
A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
What is the airplanes resultant speed (magnitude of vector)
What is the airplanes heading (direction of velocity vector
Answer:
1.) 19.21 m/s
2.) 57 degree
Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
The airplanes resultant speed can be calculated by using pythagorean theorem
R = sqrt ( 15^2 + 12^2 )
R = sqrt( 225 + 144 )
R = sqrt( 369 )
R = 19.21 m/s
The magnitude of the resultant vector is 19.21 m/s
The direction of velocity vector will be:
Tan Ø = 15 /12
Tan Ø = 1.25
Ø = tan^-1(1.25)
Ø = 57.04
Ø = 57 degree.
Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector
Explain How can you
use a distance-time graph
to calculate average
speed?
Explanation:
Average speed = change in distance / change in time
In other words, the average speed is the slope of the line connecting two points on the graph.
NEED HELP FAST!!
As the air on the surface of the Earth warms, what happens to the density of the air?
A.It decreases
B.It increases
C.It remains constant
D.It decreases, then increase
Answer:
B
Explanation:
because when the air rises the density increases
We want to predict what will happen to the density of the air on the surface of the Earth when it warms up.
We will see that the correct option is D: "It decreases, then increases"
We know that the temperature of a given object (in this case a mass of air) is related to the kinetic energy of the particles that conform it.
As the temperature increases, the kinetic energy also increases, thus, the amount of motion of each particle increases, thus, the volume of the object increases.
Now remember that:
density = Mass/Volume.
So if the volume of something increases, we will see that the density decreases (as the volume is in the denominator).
Then if the temperature of the air increases, we will see that the density of the air in the surface decreases.
But it does not end there, as only the air near the surface suffers this change of density, we will have a denser mass of air (colder air) above it. And because it is denser (has more mass in the same volume) we can say that it is heavier.
Then eventually the hot air will rise, and the cold air will fall down, thus the density of the air in the surface increases again, as the colder and denser air comes near the surface.
This is one way of how wind currents are born.
Concluding we can see that the correct option is D: "It decreases, then increases"
If you want to learn more, you can read:
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You must exert a force of 5N on a book to slide it across a table. If you do 2.5 J of work in the process, how far has the book moved?
Answer:
0.5 m
Explanation:
The following data were obtained from the question:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:
Work done (W) = Force (F) × Distance (s)
W = F × s
With the above formula, we can obtain the distance moved by the book as shown below:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
W = F × s
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
Therefore, the book moved a distance of 0.5 m.
At a time of 30 seconds a runner passes a distance marker labeled "125 meters." If the velocity of the runner is +5.0 m/s, when did the runner pass the distance marker for 75 meters?
Answer:
Explanation
He runs at 5m/s, so in 30 s he should be at 150m. So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.
When a wave goes through a substance, _______________ has occurred.
When a wave gets trapped or stuck in a substance, ______________ has occurred.
When a wave bounces off of a surface, ______________ has occurred.
The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
Answer:
A) When a wave goes through a substance, _______________ has occurred.
This is refraction or transmission.
Refraction refers to the changes in the wave that occur when it enters a new medium, in this case, the substance.
Transmission refers to the transfer of energy that is now in the medium.
Usually, we have both of those together.
B) When a wave gets trapped or stuck in a substance, ______________ has occurred.
This is total internal reflection. This happens when, by the Snel Law, the angle of the transmited wave is 90° (or larger) which is traduced in no transmitted wave, this is why the wave is "trapped" in the substance.
C) When a wave bounces off of a surface, ______________ has occurred.
This is reflection.
D) The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
This is the Law of Reflection, that says "when light is reflected from a surface, the angle of reflection is the same as the angle of incidence"
Thermal effects refers to the:
Answer:
removal of heat by cooling towers
If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the magnitude of the resultant velocity?
Answer:
40 m/s
Explanation:
How many elements in oxygen gas? PLEASE ANSWER!
Answer:
8
Explanation:
Answer:
Chemical Properties of Oxygen
At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.
Explanation:
A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a with negative charge −e distributed uniformly around the ring (representing the electron in orbit around the proton). Find the magnitude of the total electric field due to this charge distribution at a point a distance a from the proton and perpendicular to the plane of the ring.
Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
4. What is the momentum of a 78.0 N bowling ball with a velocity of 8.00 m/s?
Momentum = (mass) x (speed)
We don't know the mass of the bowling ball, but we know how much it weighs. So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass. The question doesn't tell us what planet the bowling ball is on. So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.
But I urgently need the points, so I'm going to go out on a limb here and make a big assumption: I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration of gravity is around 9.8 m/s² everywhere on the planet's surface. NOW I can go ahead and answer the question that I just invented.
weight = m g
Mass = (weight) / (g)
Mass = (78.0 N) / (9.8 m/s²)
Mass = 7.96 kilograms
Momentum = (mass) x (speed)
Momentum = (7.96 kg) x (8.00 m/s)
Momentum = 63.7 kg-m/s
The bird that migrates the farthest is the Arctic tern. Each year, the Arctic tern travels
32,000 km between the Arctic Ocean and the continent of Antarctica. Most of the
migration takes place within two four-month periods each year. Assume an Arctic
tern completes the second half of its annual migration distance in 122 days. Also
assume that during this time the tern flies directly north. If the tern flies the same
distance each day, what is its velocity in kilometers per day?
Answer: 131.14km per day
Explanation: since the second half of the terns migration takes 122 days we can assume that the full migration would take 244 days. using this we can divide the total distance by the total amount of days it takes (because speed = distance/time) which is 32,000/244, which would be 131.14
which is an example of a vector quantity
A.time
B.speed
C. acceleration
D. distance
Answer:
The answer to your question is C
Explanation:
acceleration is a vector quantity because it has both magnitude and direction
What is the correct answer choice to the question above?
Answer: A. a straight line inclined to the time axis
Explanation:
g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
W= 0.319992 J and distance is 3.75 cm
Explanation:
The energy needed to stretch the spring from 30 cm to 45 cm = 3 J
Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.
So first find the work done to stretch the spring from 35 cm to 45 cm.
Work done, w = (1/2) kx^2
3 = (1/2)k(0.45 – 0.30)^2
k = 266.66 N/m
now, x1 = 0.35 – 0.30 = 0.05m
x2 = 0.37 – 0.30 = 0.07m
Now the amount of work done to stretch from 35cm to 37.
w = (1/2) k (x2^2 – x1^2)
w = (1/2) (266.66) (0.07^2 – 0.05^2 )
w= 0.319992 J
(b). Given F = 10 N
F = kx
x = F / k
x = 10 / 266.66
x = 0.0375m
x = 3.75 cm
Thus, distance is 3.75cm
three short square wood blocks measuring 3.5 per side support a machine weighing 29500 lbs. What is the compressive stress in the blocks
A. 798lbs
B. 421lbs
C. 1404lbs
D. 803lb
Answer:
D. 803 lbs
Explanation:
In order to find the compressive stress on all three blocks we first need to find the normal surface area of each:
Surface Area of 1 Block = 3.5 x 3.5
Surface Area of 1 Block = 12.25
Surface Area of all 3 Blocks = A = 3 x 12.25
Area = 36.75
Now, the stress is given by the following formula:
Stress = Force/Area
Stress = 29500 lbs/36.75
Stress = 802.72 lbs
Hence, the correct option will be:
D. 803 lbs
HELP!!!
An arrow is shot into the air at an angle of 30.0 above the horizontal with a speed of 20.0 m/s. What are the x and y components of the
velocity of the arrow 1.0 s after it leaves the bowstring?
Answer:
Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1
X(1s) = 10m/s
Explanation:
In annex I've done the explanation for the equations that I will just present here.
Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:
[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]
Ignoring friction with air, Xo = Xf
So, Xo is the same during all the movement.
X(1s) = 10m/s
For Yo is different. That component is suferring reductions from gravity.
We can find Yo(1s) with one the basic functions of cinematics:
Vf = Vo + at
Vf = Final Velocity
Vo = Start Velocity
a = aceleration - gravity (g) is negative here
t = time
Yf = Yo + gt
Yf = [tex]10\sqrt{3}[/tex] - 10.1
If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])
A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]
Answer:
90 m
Explanation:
There are two unknowns: the amount of time the squirrel spent running, and the length of the race. Let's call these t and x, respectively.
The average velocity is the total distance divided by the total time.
5.0 m/s = x / (t + 3.0)
The total distance is the time she spent running times the speed she ran at.
x = (6.0 m/s) t
Substitute and solve:
5 = 6t / (t + 3)
5 (t + 3) = 6t
5t + 15 = 6t
t = 15
She ran for 15 seconds (not including the 3 seconds she stopped). So the length of the race is:
x = (6.0 m/s) (15 s)
x = 90 m
if AN OBJECT HAS ZERO NET FORCE ON IT THEN THE OBJECT MUST BE AT REST true or false
Answer:
False
Explanation:
constant velocity is also 0 net force
Answer:
FALSE
Explanation:
The object can be at rest or also could be in uniform motion (moving with constant velocity)
A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed? Take the speed of sound to be 343 m/s.
Answer:
The value is [tex]v_s = 1.394 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]
The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s [/tex]
Generally the frequency of the note played by the first player is mathematically represented as
[tex]f_1 = f_2 + f_b[/tex]
=> [tex]f_1 = 490 + 2.0 [/tex]
=> [tex]f_1 = 492 Hz[/tex]
From the relation of Doppler Shift we have that
[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]
Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]
So
[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]
=> [tex]v_s = 1.394 \ m/s[/tex]
A statement about what happens in nature that seems to be true all the time ES
Answer:
their is always a animal or bug in nature
Explanation:
Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temperature), and measured terminal speed of 2.64X10^-5 m/s and weight-neutralizing voltage of 35.0 V. How many electrons have been attached to, or detached from the initially neutral plastic sphere
Following are the calculation to the given question:
Given:
Please find the given question.
To find:
value=?
Solution:
With terminal voltage, no force equals zero.
So,
[tex]\to mg = 6\times \pi \times \eta \times r\times \vartheta \\\\[/tex]
As [tex]35\ V[/tex] is weight neutralizing voltage
[tex]\to mg= q\times v \times r\\\\[/tex]
So,
[tex]\to q\times v \times r= 6 \times \pi \times \eta \times r \times \vartheta \\\\\to q\times v = 6 \times \pi \times \eta \times \vartheta \\\\\to q =\frac{6 \times \pi \times \eta \times \vartheta}{v} \\\\[/tex]
[tex]=\frac{6 \times 3.14 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 18.84 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\ =3.2 \times 10^{-10}\ coulombs[/tex]
Therefore
[tex]\to q= n \times e\\\\[/tex]
Hence n comes to be [tex]2\times 10^9\[/tex] electrons.
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The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 kilo-newtons and the space shuttle has an acceleration of 16,000 miles per hour squared , what is the mass of the spacecraft in units of pounds-mass ?
Answer:
m = 81281.5 pounds.
Explanation:
Given that,
Force, F = 73 kN
Acceleration of the space shuttle, a = 16000 mi/h²
1 miles/h² = 0.0001241 m/s2
16000 mi/h² = 1.98 m/s²
We need to find the mass of the spacecraft.
According to Newton's second law,
F = ma
m is mass of the spacecraft
[tex]m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg[/tex]
Since, 1 kg = 2.20462 pounds
m = 81281.5 pounds
Hence, the mass of the spacecraft is 81281.5 pounds.
What happens if we increase the value of the resistor in forward bias connection?
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.273.27 times a second. A tack is stuck in the tire at a distance of 0.365 m0.365 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/sm/s What is the tack's centripetal acceleration?
Answer:
The tangential speed is [tex]v = 7.5 m/s [/tex]
The centripetal acceleration is [tex]a = 154 \ m/s^2[/tex]
Explanation
Generally the angular velocity is mathematically represented as
[tex]w = e * \frac{ 2 \ pi \ rad }{s}[/tex]
substituting 3.27 rev/s for e we have that
[tex]w = 3.27 * \frac{ 2 \ pi \ rad }{s}[/tex]
=> [tex]w = 20.55 \ rad /s[/tex]
The tangential speed is mathematically represented as
[tex]v = w * r[/tex]
substituting 0.365 m for r we have that
[tex]v = 20.55 * 0.365 [/tex]
=> [tex]v = 7.5 m/s [/tex]
Generally the centripetal acceleration is mathematically represented as
[tex]a = \frac{v^2}{r}[/tex]
=> [tex]a = \frac{7.5^2}{ 0.365}[/tex]
=> [tex]a = 154 \ m/s^2[/tex]
Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?
Answer:
Explanation:
we know that
s=vt here v is the speed and s is distance covered by the signals
given data
v=3*10^8
t=10 min we have to convert it into seconds
1 minute=60 seconds
so
10 minutes =10*60/1 =600 seconds
now putting the value of v and t we can find the value of s
s=vt
s=3*10^8*600
s=1.8*10^11m
i hope this will help you
Speed is the rate of distance over time.
The signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes
The given parameters are:
[tex]s = 3 \times 10^8\ ms^{-1}[/tex] ---the speed of light
[tex]t = 10\ min[/tex] -- the time of travel
The relationship between speed, distance (d) and time is:
[tex]s = \frac dt[/tex]
Make d the subject
[tex]d = s \times t[/tex]
Substitute values for t and s
[tex]d = 3 \times 10^8\ ms^{-1} \times 10\ min[/tex]
Convert minutes to seconds
[tex]d = 3 \times 10^8\ ms^{-1} \times 10 \times 60s[/tex]
So, we have:
[tex]d = 3 \times 10^8\ m \times 10 \times 60[/tex]
[tex]d = 18 \times 10^{10}m[/tex]
Rewrite as:
[tex]d = 1.8 \times 10^{11}m[/tex]
Hence, the signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes
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I'm confused, and I can't seem to get this question right. someone please help. Only two forces act on an object (mass = 5.00 kg), as in the drawing. (F = 55.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. (Drawing: 45 degrees, Fx is 40 N)
Answer:
17.6 m/s², 26.2° above x axis
Explanation:
Apply Newton's second law.
Sum of forces in the x direction:
∑Fₓ = ma
40 N + 55.0 N cos 45° = (5.00 kg) aₓ
aₓ = 15.8 m/s²
Sum of forces in the y direction:
∑Fᵧ = ma
55.0 N sin 45° = (5.00 kg) aᵧ
aᵧ = 7.78 m/s²
Use Pythagorean theorem to find the magnitude of the resultant.
a² = aₓ² + aᵧ²
a² = (15.8 m/s²)² + (7.78 m/s²)²
a = 17.6 m/s²
Use trigonometry to find the angle.
tan θ = aᵧ / aₓ
tan θ = (7.78 m/s²) / (15.8 m/s²)
θ = 26.2°