Please help really easy

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Answer 1

Answer:

Answer:

I am sure it is A because no chemical change occurs and it is a physical change. If you can Brainllest than that would be great but if you wanna you don't have to. Hope this helps!! If wrong sorry.

Explanation:

Related Questions

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.
What it the Zeeman effect and what does it tell us about the Sun?
A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom? b. Express your answer to two significant figures and include the appropriate units. c. How far does the water move horizontally during this time? d. Express your answer to two significant figures and include the appropriate units.
Patricia places 1500g of fruit on a scale compressing it by 0.02m. What is the force constant of the spring on the scale. Hint: you need 2 equations to solve this.

A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?

Answers

The work required to bring the sphere to stop is equal to the kinetic energy possessed by the sphere.

Kinetic energy of a rotating body is given by,

K.E = $(1)/(2)Iw^(2)$

Here, I= Moment of inertia of hollow sphere,

Since, the hollow sphere is rotating about the axis passing through its center, I = $(2)/(3)MR^(2)$

M= Mass of the sphere= 3.8 kg,

R= Radius of gyration= Radius of the sphere= 0.25 m

w= Angular speed of the sphere = 13 rad/s

Substituting the values,

Kinetic energy = $(1)/(2) *(2)/(3) (3.8)(0.25)^(2)(13.0)^(2)$

= 13.4 J

∴ Work required to bring the sphere to stop is 13.4 J.

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?

Answers

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

$\alpha = (w2^(2)-w1^(2) )/(2*(\theta2 - \theta1))$

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²

t=1.097 sec

A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 x 103N with an effective perpendicular arm of 3.0 cm, producing an angular acceleration of the forearm of 130 rad/s2

What is the moment of inertia of the boxer's forearm?

Answers

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

$\tau = F * r$

Where,

F = Force

r = Distance

Replacing,

$\tau = 2*10^3*0.03$

$\tau = 60N\cdot m$

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

$\tau = I \alpha$

I = Moment of inertia

$\alpha$ = Angular acceleration

Replacing with our values we have that

$I = (\tau)/(\alpha)$

$I = (60)/(120)$

$I = 0.5kg\cdot m^2$

Therefore the value of moment of inertia is $0.5kg\cdot m^2$

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + (1)/(2)(-g)t^2$

=> $-532 = 71.83 t + (1)/(2)(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^(-1)[(v_y)/(v_x ) ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^(-1)[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

Calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Answers

Answer:

For 25-turn electromagnet, Number of clips = 4.1

For 50-turn electromagnet number of clips = 9.6

Explanation:

To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Hence;

Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get

For 25-turn electromagnet y = 3.663x * 0.5

(rounded to one decimal place) Number of clips = 4.1

For 50-turn electromagnet y = 7.133x 2.5

(rounded to one decimal place) Number of clips = 9.6

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Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.

An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? yr (b) How far is the probe from Earth when its batteries fail, as measured by mission control? ly (c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail? ly

A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?