A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s.

Required:
a. Compute the magnitude and direction of the velocity of the stone after it is struck.
b. Is the collision perfectly elastic?

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

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

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.

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.

Answer:

B. 15 miles an hour going west

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]

Answer: Hazmat products are allowed in your FC are:

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.

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:

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

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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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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Explanation:

everything can be found in the picture

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

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.

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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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.

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

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

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,

Mass,

Block's coefficient,

By 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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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.

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

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

Answer:

171.5J

Explanation:

K=1/2 *m *U²

K=1/2 *7 *7²

K=171.5 J

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.)

Answer:

So as KE becomes 4times , Momentum will increase by 2 times.

Answer:

150000 is the answer to the question

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

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.

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