PHYSICS COLLEGE

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.

Answers

Answer 1

Answer:

Answer :

a.3.29 m/s

b.3.3 m/s

c.14.8 m

d.15 m

Explanation:

We are given that

Initial Horizontal speed= $v_x=4.5 m/s$

Vertical component of initial speed= $v_y=0$

Vertical distance= $y=-53 m$

a. $s=ut+(1)/(2)gt^2$

Using the formula and g is negative therefore $g=-9.8m/s^2$

$-53=0-(1)/(2)(9.8)t^2$

$53=4.9t^2$

$t^2=(53)/(4.9) s$

$t=\sqrt{(53)/(4.9)}=3.29m/s$

b.Hundredth place is greater than 5 therefore, 1 will be added to tenth place and other digits on left side of tenth place remains same and digit on right side of tenth place replace by 0.

$t=3.3 m/s$

c.Horizontal acceleration= $a_x=0$

$x=v_xt=4.5* 3.29=14.8 m$

d.Tenth place 8 is greater than 5 therefore, 1 will be added to unit place and other digits on left side of unit place remains same and digit on right side of unit place replace by 0.

Horizontal distance=15 m

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The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

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Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?

Answers

Answer:

$N_T=2086285.67$

Explanation:

Given;

Thickness of the glass plate, $x=2.95* 10^(-3)\ m$

refractive index of the glass plate, $n=1.6$

wavelength of light source in vacuum, $\lambda=600* 10^(-9)\ m$

distance between the source and the screen, $d=1.25\ m$

Distance travelled by the light from source to screen in vacuum:

$d_v=d-x$

$d_v=1.25-0.00295$

$d_v=1.24705\ m$

So the no. of wavelengths in the vacuum:

$N=(d_v)/(\lambda)$

$N=(1.24705)/(6* 10^(-7))$

$N\approx2.0784* 10^(6)$ .......................(1)

Now we find the wavelength of the light wave in the glass:

$n=(\lambda)/(\lambda')$

where:

$\lambda'=$ wavelength of light in the medium of glass.

$1.6=(600* 10^(-9))/(\lambda')$

$\lambda'=375* 10^(-9)\ m=375\ nm$

Now the no. of wavelengths in the glass:

$N'=(x)/(\lambda')$

$N'=(2.95* 10^(-3))/(375* 10^(-9))$

$N'=7.8667* 10^(3)$ ............................(2)

From (1) & (2):

total no. of wavelengths are there between the source and the screen:

$N_T=N+N'$

$N_T=2086285.67$

Which two types of energy does a book have as it falls to the floor

Answers

Answer:

kinetic and potential energy

Explanation:

It's a little hard can u help

Answers

I can't really tell what (I) is, but F) blender def converts electric into motion of spinning blades

You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

Answers

Answer:

μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

a = 2 x / t²

a = 2 0.5 /0.5²

a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

X axis

fr - Wₓ = m a (1)

Y Axis

N- $W_(y)$ = 0

N = W_{y}

We use trigonometry to find the components of the weight

sin 45 = Wₓ / W

cos 45 = W_{y} / W

Wₓ = W sin 45

W_{y} = W cos 45

The out of touch has the expression

fr = μ N

fr = μ W_{y}

We substitute in 1

μ mg cos 45 - mg sin 45 = m a

W_{y} = (a + g sin 45) / g cos 45

μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

a = -4 m / s²

μ = 1- 4 / (9.8 cos 45)

μ = 0.423

Answer:

The μ = 0.422

Explanation:

The distance travelled by the mass is equal to:

$L=ut+(1)/(2)at^(2) \n0.5=(0*5)+(1)/(2) a(0.5^(2) )\na=4m/s^(2)$

The sum of forces in y-direction equals zero:

∑Fy = 0

N - (m * g * cosθ) = 0

N - (1 * 9.8 * cos45) = 0

N = 6.93 N

The sum of forces in x-direction is equal to:

∑Fx = ma

(m * g * sinθ) - fk = m * a

(1 * 9.8 * sin45) - fk = 1 * 4

fk = 2.93 N

fk = μ * N

2.93 = μ * 6.93

μ = 0.422

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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25 POINTS FIRST CORRECT GET BRAINLIEST!!!!!!!!!!!!!!!!1