A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero

(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero


(c) Find the initial velocity of the ball.

______m/s upward

(d) Find the maximum height it reaches.
___________m

The coefficient of friction is 1.39.

Potential energy is a form of energy that an object possesses due to its position or configuration. It is the energy that is stored in an object as a result of its position or shape, and is potentially available to do work when released. The amount of potential energy an object has depends on its mass, height, and its distance from a reference point where the potential energy is considered to be zero. The most common example of potential energy is gravitational potential energy, which is the energy an object has due to its position above the ground. Other examples of potential energy include elastic potential energy, chemical potential energy, and nuclear potential energy. Potential energy can be converted into other forms of energy, such as kinetic energy or thermal energy, through various processes such as falling, stretching, or chemical reactions.

First, let's find the spring potential energy when it is compressed by 10.0 cm:

U_spring = (1/2)kx² = (1/2)(25.0 N/m)(0.1 m)² = 0.125 J

When the object is released, all of this potential energy is converted to kinetic energy:

K = U_spring = 0.125 J

The initial velocity of the object can be found using the conservation of energy:

K = (1/2)mv²

v = √(2K/m) = √(2(0.125 J)/(0.0200 kg)) = 2.50 m/s

Since the object slides a distance of 1.15 m, the average frictional force acting on it is:

F_friction = (1/2)mv²/d1 = (1/2)(0.0200 kg)(2.50 m/s)²/1.15 m = 0.272 N

The gravitational potential energy of the object when it is at a height h = 1.00 m above the floor is:

U_gravity = mgh = (0.0200 kg)(9.81 m/s²)(1.00 m) = 0.196 J

When the object falls to the floor, this potential energy is converted to kinetic energy:

K = U_gravity = 0.196 J

The final velocity of the object just before it hits the floor can be found using the conservation of energy:

K = (1/2)mv²

v = √(2K/m) = √(2(0.196 J)/(0.0200 kg)) = 3.14 m/s

The horizontal distance the object travels while falling can be found using the time it takes to fall:

t = √(2h/g) = √(2(1.00 m)/(9.81 m/s²)) = 0.452 s

d_fall = v*t = (3.14 m/s)(0.452 s) = 1.42 m

The total horizontal distance the object travels is:

d_total = d1 + d_fall = 1.15 m + 1.42 m = 2.57 m

The coefficient of friction can be found using the relationship:

F_friction = μN

where N is the normal force acting on the object. Since the object is not accelerating vertically, we know that:

N = mg

The normal force is also equal in magnitude to the force that the tabletop exerts on the object perpendicular to the surface:

N = F_tabletop

Therefore, the coefficient of friction is:

μ = F_friction/F_tabletop = F_friction/mg = (0.272 N)/(0.0200 kg)(9.81 m/s²) = 1.39

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Answer:

Forwarded

Explanation:

Answer:

kinetic

Explanation:

i just remember it from last year

Answer:

kinetic energy

Explanation:

expression for kinetic energy is

kinetic energy = (1/2) × mass × (velocity)^2

so , as velocity increases K.E increases

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \(\frac{0. 06}{0. 12* 1* 0. 9}\)

friction factor = 0. 52 × \(\frac{0. 06}{0. 108}\)

friction factor = 0. 52 × 0. 55

friction factor \(= 0. 289\)

b. When V = 3mls

Friction factor = 0. 52 × \(\frac{0. 06}{0. 12 * 3* 0. 9}\)

Friction factor = 0. 52 × \(\frac{0. 06}{0. 324}\)

Friction factor = 0. 52 × 0. 185

Friction factor \(= 0.096\)

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \(\frac{1}{2*6. 6743 * 10^-11}\) × \(\frac{1^2}{0. 120}\) × \(\frac{1}{100}\)

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = \(0. 096\) × \(\frac{1}{1. 334 *10^-10}\) × \(\frac{3^2}{0. 120}\) × \(\frac{1}{100}\)

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

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To solve this problem, we need to determine the frictional force acting on the block with different magnitudes of the applied force.

First, we need to find the normal force on the block, which is equal to the weight of the block. The weight of the block is given by:

W = mg = 2.5 kg x 9.8 m/s^2 = 24.5 N

Next, we need to find the force of the applied vertical force, which is given in the problem as "is". We can use trigonometry to find the vertical component of the force:

Fv = is sinθ

where θ is the angle between the force and the horizontal surface. Since the problem does not give us the value of θ, we will assume it to be 0°, which means the force is purely horizontal.

(a) If the magnitude of the applied force is 8.0 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 0) = 9.8 N

(b) If the magnitude of the applied force is 10 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 10) = 5.8 N

(c) If the magnitude of the applied force is 12 N, then the frictional force can be calculated as:

Ff = μkFn = μk(mg - Fv) = 0.25(24.5 - 12) = 3.1 N

Therefore, the magnitude of the frictional force acting on the block is 9.8 N, 5.8 N, and 3.1 N, for applied forces of 8.0 N, 10 N, and 12 N, respectively.

(a) When the horizontal force is 8 N the frictional force is 11.8 N.

(b) when the applied force is 10 N; the frictional force is 13.8 N.

(c) when the applied force is 12 N; the frictional force is 15.8 N.

(a) The magnitude of the frictional force on the block when the horizontal force is 8 N is calculated as;

F - Ff = ma

where;

F - μmg = ma

6 - 0.4 x 2.5 x 9.8 = 2.5 a

2.5 a = -3.8

a = -3.8/2.5

a = -1.52 m/s²

when the applied force is 8 N;

8 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 11.8 N

(b) when the applied force is 10 N;

10 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 13.8 N

(c) when the applied force is 12 N;

12 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 15.8 N

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer: 1.8J

Explanation:

• Formula for Elastic Potential Energy: PEelastic = 1/2kx^2

• Convert cm to m: (30/100) = 0.30m

• PEelastic = (1/2)(40N/m)(.30m)^2

= 1.8J

Answer:

The charge is believed to be from the charge of the quarks that make up the nucleons (protons and neutrons). A proton is made of two Up quarks, with 2/3 positive charge each and one Down Quark with a negative 1/3 charge (2/3 + 2/3 + -1/3 = 1).

Explanation:

Hope this helps :)

Explanation:

The laser cauterizes as it cuts thus reducing blood loss.

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

basketball

Explanation: cause it is

Answer:

It honestly depends on which one you're better at.

Explanation: I mean that's what I think.

Answer:

hi the question isn't obvious and need a photo I guess

Answer:

oooop tehehehehehee

ahahhhahaaahahjabasa

Answer:

uhm- heh, wut lol ️️

Answer:

Curran is testing out a newly waxed hallway floor by sliding in his socks.

He runs from one end of the hallway and starts sliding midway going all

the way to the right. The forces acting on Curran as he is sliding are

represented in the diagram below. Identify the forces and match them

correctly to the n

004) The force F exerted by each hailstone is given by Second Newton law:

where m is the mass of each hailstone and a is its acceleration. a can be calculated by using:

where v is the speed at which the hailstones impact the window and t the time interval of the impact.

By replacing the values of v and t you obtain for acceleration:

Then, the force F is:

where the mass m is used in kg in order to obtain N units for force.

Now, consider that the impact of the hailstones on the window is at angle of 44 degrees. It means that the effective force exerted by each hailstones is:

It means that the average force due to the total number of hailstones is:

005) Then, by replacing the previous value of the force and the area of the impact on the window, into the formula for the pressure, you have:

Answer:

The direction could change

Explanation:

Sooooo, first,

Constant Speed means that the body isn't accelerating.

a= 0

Let the force that causes it to move up the incline be P and Q be the force that causes the body to move down the incline. Let Fr be the frictional force.

The force that pulls the body downwards is given by,

F = mgsin40°

F= 18.898N

P -(F+Fr) = ma, but, a= 0

P = F + Fr

Fr = 26-18.898

= 7.102N

If the block must move down the incline, Q and Fr will act in the same direction.

Q + Fr = F

Q = 11.796N

1. The electric field pattern due to charged sphere A can be represented by lines radiating outward from the sphere.

2. The magnitude of the charge on sphere A is approximately 0.0833 Coulombs.

3. The electrostatic force experienced by sphere B when placed at point P is approximately 2.675 x 10^-4 Newtons.

1. These lines should be evenly spaced and symmetric around the sphere, indicating a radial field pattern. Since the electric field at point P is directed towards sphere A, the field lines should point inward towards the sphere. Thus, the electric field pattern would resemble a series of concentric circles with lines converging towards the center of sphere A.

2. To calculate the magnitude of the charge on sphere A, we can use the formula for the electric field strength (E) due to a point charge:

E = k * (Q / r^2)

where k is the electrostatic constant (approximately 9 x 10^9 N m^2/C^2), Q is the charge on the sphere, and r is the distance from the sphere to the point P.

From the given information, we have E = 3 x 10^7 N/C and r = 0.5 m. Plugging these values into the formula and solving for Q:

3 x 10^7 N/C = (9 x 10^9 N m^2/C^2) * (Q / (0.5 m)^2)

Simplifying the equation, we find:

Q = (3 x 10^7 N/C) * (0.5 m)^2 / (9 x 10^9 N m^2/C^2)

Q ≈ 0.0833 C (Coulombs)

Therefore, the magnitude of the charge on sphere A is approximately 0.0833 Coulombs.

3. When sphere E, which has an excess of 105 electrons, is placed at point P, it will experience an electrostatic force due to the interaction with sphere A. The electrostatic force between two charges can be calculated using Coulomb's law:

F = k * (|q1| * |q2|) / r^2

where k is the electrostatic constant, q1 and q2 are the charges on the spheres, and r is the distance between them.

Since each electron carries a charge of approximately -1.6 x 10^-19 C, the excess charge on sphere E is:

q2 = 105 electrons * (-1.6 x 10^-19 C/electron)

Plugging in the values and the given distance of 0.5 m, we have:

F = (9 x 10^9 N m^2/C^2) * (|0.0833 C| * |-1.6 x 10^-19 C|) / (0.5 m)^2

Simplifying the equation, we find:

F ≈ 2.675 x 10^-4 N (Newtons)

Therefore, the electrostatic force experienced by sphere E when placed at point P is approximately 2.675 x 10^-4 Newtons.

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Humans would need a breathable environment like that on Earth in the living section of a colony on Venus in order to survive. Nitrogen, oxygen, and trace amounts of other gases, such as carbon dioxide, make up the majority of the atmosphere on Earth.

The clouds are made of sulfuric acid, and the atmosphere is primarily carbon dioxide, the same gas that causes the greenhouse effect on Venus and Earth. And the heated, high-pressure carbon dioxide acts corrosively at the surface.

For instance, compared to Earth, which has 99% nitrogen and oxygen in its atmosphere, Venus and Mars both contain more than 98% carbon dioxide and nitrogen.

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Answer:

12 dozen

Explanation:

Answer:

The table is proportional.

Explanation:

You would do 21/6 to get 3.5. Then 28/8 to get 3.5. After 35/10 to get 3.5. Finally 49/14 to get 3.5. If each answer (3.5) is the same than it is proportional.

hope i helped

Answer:

What is the difference between a geostationary satellite and a non-geostationary satellite? Geostationary (GSO) satellites occupy an orbital position 36,000 km above the earth, and remain in a stationary position relative to the Earth itself.

Answer:

1.363×10^15 seconds

Explanation:

The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.

To calculate the time taken from Kepler's third Law :

T^2 = ( 4π^2/GMe ) r^3

Where Me is the mass of the earth

r is the average distance travel

G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2

π = 3.14

Me = mass of earth = 5.972×10^24kg

r =( r minimum + r maximum)/2 ......1

rmin = 6590km

rmax = 385000km

From equation 1

r = (6590+385000)/2

r = 391590/2

r = 195795km

From T^2 = ( 4π^2/GMe ) r^3

T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3

= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15

= 39.4384/ 3.983×10^14 ) × 7.5059×10^15

= (9.901×10^14) × 7.5059×10^15

T^2 = 7.4321× 10^30

T =√7.4321× 10^30

T = 2.726×10^15 seconds

The time for one way trip from Earth to the moon is :

∆T = T/2

= 2.726×10^15 /2

= 1.363×10^15 secs

Let \(a\) be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

\(1.3mg - T = 1.3ma\)

where \(T\) is the magnitude of tension in the pulley cord, and

• for the mass on the right,

\(T - mg = ma\)

Eliminate \(T\) to get

\((1.3mg - T) + (T - mg) = 1.3ma + ma\)

\(0.3mg = 2.3ma\)

\(\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}\)

Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity \(v\) such that

\(v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}\)

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

\(mg - T' = ma'\)

where \(T'\) and \(a'\) are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

\(T' - 0.8mg = 0.8ma'\)

Eliminate \(T'\).

\((mg - T') + (T' - 0.8mg) = ma' + 0.8ma'\)

\(0.2mg = 1.8 ma'\)

\(\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}\)

Now, the right-hand mass has an initial upward velocity of \(v\), but we're now treating down as the positive direction. As it returns to its starting position, its speed \(v'\) at that point is such that

\({v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}\)

Answer:

True

Explanation:

The time elapsed between the first splash and the second splash is approximately 0.69 seconds.

To calculate this, we consider the motion of two rocks thrown simultaneously from a bridge. Heather throws a rock straight down with a speed of 14 m/s, while Jerry throws a rock straight up with the same speed.

We use the equation for displacement in uniformly accelerated motion: s = ut + (1/2)at^2.

For Heather's rock, which is thrown downwards, the initial velocity (u) is positive and the acceleration (a) due to gravity is negative (-9.8 m/s^2). The displacement (s) is the height of the bridge (46 m).

Solving the equation, we find two possible values for the time (t): t ≈ -4.91 s and t ≈ 1.91 s.

Since time cannot be negative in this context, we discard the negative value and consider t ≈ 1.91 s as the time it takes for Heather's rock to hit the water.

For Jerry's rock, thrown upwards, we use the same equation with the same initial velocity and acceleration. The displacement is also the height of the bridge, but negative.

Solving the equation, we find t ≈ -5.68 s and t ≈ 1.22 s. Again, we discard the negative value and consider t ≈ 1.22 s as the time it takes for Jerry's rock to reach its maximum height before falling back down.

To find the time difference between the first and second splash, we subtract t ≈ 1.91 s (Heather's rock) from t ≈ 1.22 s (Jerry's rock). This gives us a time difference of approximately 0.69 seconds.

Therefore, the time elapsed between the first splash and the second splash is approximately 0.69 seconds.

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The readings of the scales will have to add 2N. This comes from the fact that each scale is supporting a part of the weight of the object. Now, if the center of gravity is midway between the two scales the readings will be the same on each spring scale; assuming this is the case the reading on each scale is 1N.