Answer:
The height of the hill is 125m.
Explanation:
Since the ball is projected horizontally from the top of a hill, there is no vertical component of the velocity of projection. Therefore, so far as motion in the vertical direction is concerned, the ball is just dropped, ie its initial velocity u is merely 0 m/s. It reaches the ground after 5 seconds from the moment of projection, under the action of accerelation due to gravity.
Using,
s = u t + ½ a t². In this expression u= 0 m/s, a = 10m/s², t = 5 s. Substituting in the equation we get,
s = 0× t + ½ ×10 m/s²× 5²s²= 5× 25 m = 125m.
A jet is circling an airport control tower at a distance of 15.8 km. An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of 9.21x10-3 radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
Answer:
The Distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
=145.5m
Explanation:
The radius of the circling jet r = 15.8km = 15.8*10^3 m
the angle subtended by moon
θ = 9.21*10^-3 rad
Therefore, the distance traveled by jet is
s = rθ
= (15.8*10^3 m)(9.21*10^-3 )
= (15,800)(0.00921)
=145.5m
The position of an object of
mass 5 kg as a function of
time is giving by r = (20m/s4)t4
i + (12 m/s3)t3 j. Find the
force acting on the object as a
function of time. Express the
force in unit vectors. Hint:
Remember that Newton's
second Law relates the force
to the acceleration
Answer:
[tex]F=5(240t^2i+72tj)\ N[/tex]
Explanation:
Given that,
The mass of the object, m = 5 kg
The position vector is, [tex]r=20t^4i+12t^3j[/tex]
Velocity, [tex]v=\dfrac{dr}{dt}=80t^3i+36t^2j[/tex]
Acceleration, [tex]a=\dfrac{dv}{dt}=240t^2i+72tj[/tex]
Newton's second law of motion is given as follows:
F = ma
Put all the values,
[tex]F=5(240t^2i+72tj)\ N[/tex]
Hence, this is the required solution.
Which best describes velocity?
A. 15 miles an hour
B. 15 miles an hour going west.
C. Speeding up from 10 to 15 miles an hour.
D. Going 15 miles an hour for 2 hours.
Answer:
B. 15 miles an hour going west
A(n) 60.9 kg astronaut becomes separated
from the shuttle, while on a space walk. She
finds herself 36.5 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0.928 kg camera in her
hand and decides to get back to the shuttle
by throwing the camera at a speed of 12 m/s
in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
Answer in units of min.
Answer:
m v = M V conservation of momentum after throwing camera
V = m * v / M = .928 * 12 / 60.9 = .183 m/s
t = S / V = 36.5 / .183 = 199 sec = 3.32 min
If you wanted to know how much the temperature of a particular piece of material would rise when a known amount of heat was added to it, which of the following quantities would be most helpful to know?
a. coefficient of linear expansion
b. specific heat
c. initial temperature
d. thermal conductivity
e. density
Answer:
Option (b) is correct.
Explanation:
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C, is called specific heat of the substance.
The formula of the specific heat is
H = m c (T' - T)
where, m is the mass, c is the specific heat and T' - T is the change in temperature.
So, to know the rise in temperature, by adding the known amount of heat, the specific heat is required.
So, option (b) is correct.
The red light from a helium-neon laser has a wavelength of 644.6 nm in air. Find the speed, wavelength, and frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)speed (m/s)wavelength (nm)frequency (Hz)airwaterglass
Answer:
air f = 4.6527 10¹⁴ Hz
water f = 3.4914 10¹⁴ Hz
glass f = 3.1027 10¹⁴ Hz
Explanation:
The refractive index of a material is given by
n = c / v
where c is the speed of light in a vacuum c = 3 108 m / s and v is the speed of light in the material medium.
the speed of the wave is
v = λ f
we substitute
c / n = λ f
f = [tex]\frac{c}{n \ \lambda}[/tex]
The refractive indices are
air 1,00029
water 1.3330
glass 1.5
let's calculate the frequencies
vaccum
f = 3 10⁸ / 1 644.6 10⁻⁹
f = 4.6540 10¹⁴ Hz
air
f = 3 10⁸ / 1,00029 644.6 10⁻⁹
f = 4.6527 10¹⁴ Hz
Water
f = 3 10⁸ / 1.333 644.6 10⁻⁹
f = 3.4914 10¹⁴ Hz
glass
f = 3 10 ^ 8 / 1.5 644.6 10⁻⁹
f = 3.1027 10¹⁴ Hz
The engine in the car from question 1 uses a force of 1200 N to cause the car to accelerate at 3.5 m/s2. What is the car's mass?
Answer:
Mass of car's engine = 342.85 kg (Approx.)
Explanation:
Given values:
Force applied by car = 1,200 N
Acceleration of car' engine = 3.5 m/s²
Find:
Mass of car's engine
Computation:
⇒ Mass = Force / Acceleration
⇒ Mass of car's engine = Force applied by car / Acceleration of car
⇒ Mass of car's engine = 1,200 / 3.5
⇒ Mass of car's engine = 342.85 kg (Approx.)
A person rolls a 7 kg bowling ball down a lane in a bowling alley. The lane is
18 m long. The ball is traveling at 7 m/s when it leaves the person's hand.
What is the ball's kinetic energy at this point?
Answer:
171.5J
Explanation:
K=1/2 *m *U²
K=1/2 *7 *7²
K=171.5 J
A kettle operates from a 120 V outlet. It has a heating element with a resistance of 8.0 Ω . Calculate the current going through the element.
Answer:
I = 15A
Explanation:
V = I*R
120V = I*8.0ohms
I = 120V/8.0ohms
I = 15A
Answer:
I=15A
Explanation:
you know what to do.
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Give them the brain list.
What is the change in internal energy if 70 J of heat is added to a system and
the system does 30 J of work on the surroundings. Uze al-Q-W.
O A. 40 J
O B. -40.3
O C. 100.
D. -1003
Answer:
A. 40 J
Explanation:
Given;
heat added to the system, Q = 70 J
work done by the system, W = 30 J
The change in the internal energy of the system is calculated using the first law of thermodynamic as shown below;
ΔU = Q - W
ΔU = 70 J - 30 J
ΔU = 40 J
Therefore, the change in the internal energy of the system is 40J
How bullet train and a circuit breaker work on the magnet effect of current
A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answer to two significant figures.
The spring constant, k, is
StartFraction N over m EndFraction.
Answer: 4800 N/m
Explanation:
Given
mass of rock [tex]m=2.5\ kg[/tex]
Height of cliff [tex]h=32\ m[/tex]
compression in the spring [tex]x=57\ cm[/tex]
Here, potential energy is converted into kinetic energy which in turn converts to elastic potential energy of the spring
[tex]\Rightarrow mgh=\dfrac{1}{2}kx^2\\\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\\\Rightarrow k=\dfrac{1568}{0.3249}\\\\\Rightarrow k=4826.100\approx 4800\ N/m[/tex]
Answer:
480
Explanation:
2021
2. When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased: A. 2 times B. 4 times C. 5 times D. 7 times
When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.
Hence, the correct option is B.
The force of impact in a collision is directly proportional to the square of the velocity. When the speed is doubled (increased from 35 mph to 70 mph), the force of impact will increase by a factor of four ([tex]2^2[/tex]). This is a fundamental concept in physics known as the kinetic energy formula, which is given by:
Kinetic Energy = (1/2) * mass * [tex]velocity^2[/tex]
Since kinetic energy is directly proportional to the square of velocity, doubling the speed results in four times the kinetic energy and, consequently, four times the force of impact.
Therefore, When you increase your speed from 35 mph to 70 mph, the force of impact you experience will be increased 4 times.
Hence, the correct option is B.
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You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
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A uniform electric field is oriented in the −x direction. The magnitude of the electric field is 6500 N/C. How will the equipotential surfaces associated with this electric field be oriented?
The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, what is the magnitude of the stabilizing component of the brachialis force
Answer:
[tex]XY=636N[/tex]
Explanation:
From the question we are told that:
Distance [tex]d=0.35m[/tex]
Angle [tex]\theta=32\textdegree[/tex]
Force [tex]F=750N[/tex]
Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by
[tex]XY=Fcos\theta[/tex]
[tex]XY=750cos 32\textdegree[/tex]
[tex]XY=636N[/tex]
The spring in the figure has a spring constant of 1400 N/m . It is compressed 17.0 cm , then launches a 200 g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?
Use the work-energy theorem to find the velocity of the block when it's released by the spring. The work done by the spring on the block as it's restored to equilibrium is
W = 1/2 kx ²
where k is the spring constant and x is the compression of the spring. So
W = 1/2 (1400 N/m) (0.170 m)² = 20.23 J
This is equal to the block's change in kinetic energy ∆K,
W = ∆K
and since it starts from rest, the initial K is zero, leaving us with
W = 1/2 mv ²
where m is the mass of the block and v is its speed, so that
20.23 J = 1/2 (0.200 kg) v ²
==> v ≈ 14.2 m/s
The block slides at this speed across the frictionless surface until it hits the incline which introduces friction.
First, you need to find the length of the incline. It forms a 45° angle, and the underlying 45°-45°-90° triangle has a hypotenuse of length √2 (2.0 m) ≈ 2.83 m.
Next, you need to find the total work done on the block as it slides up the incline. Use Newton's second law to examine the forces acting on the block during this phase:
• the net force acting on the block in the direction perpendicular to the incline is
∑ F = n - mg cos(45°) = 0
where n = mg cos(45°) ≈ 1.39 N is the magnitude of the normal force and mg cos(45°) ≈ 1.39 N is the perpendicular component of the block's weight;
• the net force acting on the block parallel to the surface is
∑ F = -f - mg sin(45°) = ma
where f = µn = 0.210n ≈ 0.291 N is the magnitude of kinetic friction, mg sin(45°) ≈ 1.39 N is the parallel component of the weight, and a is the acceleration of the block.
Only the parallel forces do work on the block, and this work is negative because friction and weight oppose the block's sliding up the incline. The total work done on the block is then
W = (-0.291 N - 1.39 N) (2.83 m) ≈ -4.74 J
Use the work-energy theorem again to find the block's new speed v at the top of the incline:
W = ∆K
==> -4.74 J = 1/2 (0.200 kg) v ² - 1/2 (0.200 kg) (14.2 m/s)²
==> v ≈ 12.4 m/s
And now this becomes a projectile problem. The block travels a horizontal distance x after being launched at an angle of 45° with initial speed 12.4 m/s after time t according to
x = (12.4 m/s) cos(45°) t
Its height y from the 2.0 m-high surface at time t is given by
y = (12.4 m/s) sin(45°) t - 1/2 gt ²
The block lands on the surface when y = 0, which occurs after t ≈ 1.79 s, at which point the block has covered a distance d ≈ 15.7 m.
The block sail through the air at the distance of "15.8 m"
Given:
Spring constant,
1400 N/mMass,
200 gBlock's coefficient,
0.210By using Work energy theorem, we get
→ [tex]W_{spring}+W_g+W_f = KE_f-KE_i[/tex]
By substituting the values, we get
→ [tex]\frac{1}{2}\times 1400\times (0.17)^2- (0.2\times 9.8\times 2)-(0.21\times 0.2\times \frac{9.8}{\sqrt{2} }\times \sqrt{2}\times 2 )= \frac{1}{2}\times 0.2\times V_f^2[/tex]
here,
[tex]V_f = 12.44 \ m/s[/tex]
→ [tex]d = \frac{V_f^2 Sin 2 \Theta}{g}[/tex]
[tex]= \frac{(12.44)^2 Sin 90^{\circ}}{9.8}[/tex]
[tex]= 15.8 \ m[/tex]
Thus the answer above is right.
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For the following questions, assume the wavelengths of visible light range from 380 nm to 760 nm in a vacuum. (a) What is the smallest separation (in nm) between two slits that will produce a ninth-order maximum for any visible light
Answer:
Explanation:
This is an interference exercise, which the case of constructive interference is described by the expression
d sin θ = m λ
in this case they indicate that we are in the ninth order (m = 9).
To be able to observe the pattern, the dispersion angle must be less than 90º
we substitute
sin 90 = 1
d = m lang
let's calculate
d = 9 λ
d = 9 380 10⁻⁰
d = 3.42 10⁻⁶
d2 = 9 760 10⁻⁹
d2 = 6.84 10₋⁶
Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume.
Required:
a. Find the highest temperature attained by the gas.
b. Find the lowest temperature attained by the gas.
c. Find the highest pressure attained by the gas.
d. Find the lowest pressure attained by the gas.
Answer:
a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm
Explanation:
For isothermal expansion PV = constant
So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,
So, P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
Since V₂/V₁ = 0.19,
P₂ = P₁V₁/V₂
P₂ = 1 atm (1/0.19)
P₂ = 5.26 atm
For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas
So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,
So, P₂V₂ⁿ = P₃V₃ⁿ
P₃ = P₂V₂ⁿ/V₃ⁿ
P₃ = P₂(V₂/V₃)ⁿ
Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19
1/0.19,
P₃ = P₂(V₂/V₃)ⁿ
P₃ = 5.26 atm (0.19)⁽⁵/³⁾
P₃ = 5.26 atm × 0.0628
P₃ = 0.33 atm
Using the ideal gas equation
P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)
P₃V₃/T₃ = P₄V₄/T₄
T₃ = P₃V₃T₄/P₄V₄
T₃ = (P₃/P₄)(V₃/V₄)T₂
Since V₃ = V₄ = V₁ and P₄ = P₁
V₃/V₄ = 1 and P₃/P₄ = P₃/P₁
T₃ = (P₃/P₁)(V₃/V₄)T₂
T₃ = (0.33 atm/1 atm)(1)273 K
T₃ = 90.1 K
So,
a. The highest temperature attained by the gas is T₁ = 273 K
b. The lowest temperature attained by the gas = T₃ = 90.1 K
c. The highest pressure attained by the gas is P₂ = 5.26 atm
d. The lowest pressure attained by the gas is P₃ = 0.33 atm
concave mirror daily application
Answer:
concave mirror use in headlights and torches
Explanation:
Concave mirrors are used in headlights and torches. The shaving mirrors are also concave in nature since these mirrors can produce enlarged clear images. Doctors use concave mirrors as head mirrors to have a clearer view of eyes, noses, and ears. The dental mirrors used by dentists are also concave.
How high can water be theoretically lifted by a vacuum pump at sea level?
A) less than 10.3 m
B) 10.3 m
C) more than 10.3 m
Explanation:
everything can be found in the picture
Which of these hazmat products are allowed in your FC?
Please choose all that apply.
A GPS unit (lithium batteries)
A subwoofer (magnetized materials)
A can of hairspray (flammable/aerosols)
Fireworks (explosives)
Answer: Hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)Explanation:
Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.
Hazardous material allowed in FC are as follows.
Magnetized material products like as speakers.Non-spillable battery products like toy cars.Lithium-ion battery containing products like laptops, mobile phones etc.Non-flammable aerosol.So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).
Thus, we can conclude that hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)kinetic energy of an object is quadrupled, momentum will change by what factor?
Answer:
So as KE becomes 4times , Momentum will increase by 2 times.
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
pls mark me brainlest
State two (2) examples of osmosis occurring in everyday life
Một mol khí đang ở điều kiện tiêu chuẩn thì bị nén váo một bình 5 lít. Nhiệt độ khí trong bình là 770 C. Tính áp suất khí?
Đáp án:
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
p=5,74atm
Giải thích các bước giải:
áp suất khí là:
pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm
Consider a point on a bicycle wheel as the wheel turns about a fixed axis, neither speeding up nor slowing down. Compare the linear and angular accelerations of the point.
a. Both are zero.
b. Only the angular acceleration is zero.
c. Only the linear acceleration is zero.
d. Neither is zero
Explain your choice
Answer:
c. Only the linear acceleration is zero.
Explanation:
The linear acceleration is defined as the rate of change of linear velocity. Since the bicycle is moving in the same direction, with the same speed, without speeding up or slowing down. Therefore, there will be no change in linear velocity and as a result, linear acceleration will be zero.
The angular acceleration is the rate of change of angular velocity. Since the angular velocity is changing its direction constantly. Therefore, it has a certain component of acceleration at all times called centripetal acceleration.
Therefore, the correct option is:
c. Only the linear acceleration is zero.
What is air
A. A Buchner substance
B. A compound
C. An element
D. A mixture
Air is classified as a mixture. Option D is the correct answer.
Air is a combination of different gases, primarily nitrogen (about 78%), oxygen (about 21%), and small amounts of other gases such as carbon dioxide, argon, and trace elements. These gases are not chemically bonded to each other, but rather exist together in the same space. Option D is the correct answer.
In a mixture, the substances involved retain their individual properties and can be separated by physical means. This is true for air as well. The gases in air can be separated through processes like fractional distillation or filtration. It's important to note that air also contains other components such as water vapor, dust particles, and pollutants, which can vary in concentration depending on the location and environmental conditions. These components further contribute to the complex nature of air as a mixture.
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Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 50 kg and the coefficient of static friction between the crate and the plane is . ms
Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution
answer:
P = 141.21 N
Explanation:
Given data:
Mass of crate = 50 kg
coefficient of static friction ( μ ) = 0.25
Calculate minimum horizontal force ( P ) that holds the crate from sliding
∑fx = 0
= P + Fcos θ - N*sinθ = 0
= P + 0.25N cos 30° - Nsin30° = 0
∴ P = 0.2835 N = 0
P - 0.2853 N = 0 ------- ( 1 )
∑fy = 0
- 50g + Ncosθ + Fsinθ
- 50*9.81 + Ncos30° + 0.25Nsin30°
∴ N = 494.942 N ----- ( 2 )
input 2 into 1
P - 0.2853 ( 494.942 ) = 0
P = 141.21 N
The extension produced on the wire by the application of 300 N force is 2 mm. The elastic energy stored on the wire is
Answer:
150000 is the answer to the question