A tennis player tosses a tennis ball straight up and then catches it after 1.25 s at the same height as the point of release.

a. What is the acceleration of the ball while it is in flight?
b. What is the velocity of the ball when it reaches its maximum height?
c. Find the initial velocity of the ball.
d. Find the maximum height it reaches.

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ([tex]W[/tex]), measured in newtons, is:

[tex]W = m\cdot g[/tex] (1)

Where:

[tex]m[/tex] - Mass of the block of ice, measured in kilograms.

[tex]g[/tex]  - Gravitational acceleration, measured in meters per square second.

If we know that [tex]m = 14\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the magnitudes of the weight and normal force of the block of ice are, respectively:

[tex]N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]N = W = 137.298\,N[/tex]

And the pull force is:

[tex]F_{pull} = 98\,N[/tex]

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

[tex]m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}[/tex] (2)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the block, measured in meters per second.

[tex]\Sigma F[/tex] - Horizontal net force, measured in newtons.

[tex]\Delta t[/tex] - Impact time, measured in seconds.

Now we clear the final speed in (2):

[tex]v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}[/tex]

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]m = 14\,kg[/tex], [tex]\Sigma F = 98\,N[/tex] and [tex]\Delta t = 1.40\,s[/tex], then final speed of the ice block is:

[tex]v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}[/tex]

[tex]v_{f} = 9.8\,\frac{m}{s}[/tex]

The final speed of the block of ice is 9.8 meters per second.

Answer:

Yes, field forces exist in nature.

Explanation:

A field force is a force experience due to an interaction with fields. In this case, a contact is not required before the force can be felt.

The three major field forces are: magnetic field, electric field and gravitational field. The magnetic fields are produced due to the interaction between the north and south poles of a magnet, the electric field is one from charges, while gravitational field is a force of attraction due to gravity on the earth. All these fields occur in nature, therefore making field forces to exist in nature.

Answer:

The velocity [tex]v = 1989.2 \ m/s[/tex]

Explanation:

From the question we are told that

    The wavelength is [tex]\lambda = 0.500 \ m[/tex]

     The maximum transverse speed is  [tex]v = 4.0 \ m/s[/tex]

      The maximum transverse acceleration is  [tex]a = 1.00 *10^{5} \ m/s^2[/tex]

Generally the frequency of the wave is mathematically represented as

                    [tex]f = \frac{w}{2 \pi }[/tex]

Here  w  is the angular speed which is mathematically evaluated as

      [tex]w = \frac{a}{v}[/tex]

=>    [tex]w = \frac{1.00 *10^{5}}{4}[/tex]

=>    [tex]w = 25000 \ rad/sec[/tex]

So

       [tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]

=>    [tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]

=>    [tex]f = 3978.4 \ Hz[/tex]

Gnerally the propagation speed of the wave is mathematically represented as

        [tex]v = f * \lambda[/tex]

=>      [tex]v = 3978.4 * 0.500[/tex]

=>      [tex]v = 1989.2 \ m/s[/tex]

Answer:

a) 113N

b) 0.37

Explanation:

a) Using the Newton's second law:

\sum Fx =ma

Since the crate is not moving then its acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = 0

Fm - Ff = 0.

Fm is the moving force

Ff is the frictional force

Fm = Ff

This means that the moving force is equal to the force of friction if the crate is static.

Since applied force is 113N, hence the magnitude of the static friction force will also be 113N

b) Using the formula

Ff = nR

n is the coefficient of friction

R is the reaction = mg

m is the mass of the crate = 31.2kg

g is the acceleration due to gravity = 9.8m/s²

R = 31.2 × 9.8

R = 305.76N

Recall that;

n = Ff/R

n = 113/305.76

n = 0.37

Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37

Answer:

v = 2.38 × 10³ m/s

Explanation:

Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.

Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,

mg' = mg/6

g' = g/6

Since g = 9.8 m/s²,

g'= 9.8 m/s² ÷ 6

g' = 1.63 m/s²

The escape velocity of the moon is thus  v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.

Substituting these into v, we have

v = √(2g'R)

v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)

v = √[5.663 × 10⁶ (m/s)²]

v = 2.38 × 10³ m/s

Answer:

initial velocity = 7.44 m/s   (3 s.f.)

Explanation:

a = 6.24 m/s²     t = 0.300 s     v = 9.31 m/s       u = ?

                               v = u + at

                       9.31 = u + (6.24 x 0.300)

                             9.31 = u + 1.872

                                    u = 7.438

                                    = 7.44 m/s       (3 s.f.)

Hope this helps!

Answer:

a) 0.5644 m/s

b) ~0.384 m/s^2

c) ~3.638 m/s

Explanation:

a) Tangential speed is found be the radius*rotational speed, so it is 0.83*0.68 = 0.5644 m/s

b) Centripetal acceleration is found by v^2/r, so it is (0.5644^2)/0.83 = ~0.384 m/s^2

c) Let the tangential speed be v. The maximum centripetal force 110 N (as given). Centripetal force = mass*centripetal acceleration = mass*v^2/r (because centripetal acceleration is found by v^2/r). Inputting the values from the problem and solving for v, we get:

110 = 6.9*v^2/0.83

v = sqrt(110*0.83/6.9) = ~3.638 m/s

I hope this helps! :)

Answer:

no, otherwise they just cause a side effect on our lives

Answer:

a. up

Explanation:

As per the rule of Fleming left hand, the three fingers should be places in a perpendicular manner i.e. mutually also.

The fore finger depicts the field direction

The middle finger depicts the velocity

And, the thumb finger depicts the force direction that experienced on that particle i.e. charged

So the electrons would be deflects to up

Hence, the correct option is a.

Answer:

5800

Explanation:

caluculatior

The ball was in the air for 3.625 second.

Velocity is the direction at which an object is moving and serves as a measure of the rate at which its position is changing as seen from a specific point of view and as measured by a specific unit of time (for example, 60 km/h northbound).

In kinematics, the area of classical mechanics that deals with the motion of bodies, velocity is a fundamental idea.

A physical vector quantity called velocity must have both a magnitude and a direction in order to be defined.

Given that: the a horizontal velocity of the ball: v = 40.0 m/s .

The ball  traveled 145 meters.

Hence, the time during which the ball is in the air:

= distance travelled/ horizontal velocity of the ball

= 145 meter /  40.0 m/s .

= 3.625 second.

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Answer:

B )-14000N

Explanation:

F=mv-mu

t

F =2550(0)-2550(6)

1.1

F = -13909.09

approximately -14000N

Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, the force did the pile of hay stop the tractor 14000N

Force can be a unit of  pushing or pulling of any object which result   from the object’s interaction or movement, without applying force the objects can not be moved, can be stopped or change the direction.

Force is a  quantitative parameter between two physical bodies, means an object and its environment, there are various  types of forces in nature.

If an object present in its moving state will be either static or motion, the position of the object will be changed if it is pushed or pulled and The external push or pull upon the object is mainly called as Force.

The contact force  that occurs when we apply some effort on an object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force

Non-Contact forces are the type of forces that occur from a distance  such as Electromagnetic Force, Gravitational Force, Nuclear Force

F=mv-mu

t

F =2550(0)-2550(6)

1.1

F = -13909.09

approximately -14000N

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Answer:

Should be -39.2 N

Explanation:

w=mg

w=4 x -9.8 m/s2

= -39.2 N

Answer:

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path  

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

         [tex]a_{c}[/tex] = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

              [tex]a_{BC}[/tex] > [tex]a_{CD}[/tex]

In  summary lower accelerations are

 AB = DE <CD <BC

Answer: OVERVIEW

Students will use black and white construction paper and a light source to learn that dark objects

absorb more light and reflect less light than bright objects. The activity also demonstrates the conversion

of radiant light energy into heat energy.

CONCEPTS

• Dark surfaces absorb more visible light energy than bright surfaces

• Dark surfaces reflect less visible light energy than bright surfaces

• Energy can change forms, in this case from radiant light energy to heat

• Clouds, being bright, reflect significant amounts of sunlight and help to regulate Earth’s temperature

MATERIALS

• 2 thermometers

• Flood lamp, desk lamp, or area in direct sunlight

• Ruler

• Construction paper, 1 piece white, 1 piece black, or 2 sheets photocopy paper

• Scissors

• Cellophane tape or rubber bands

• 2 empty metal food cans, same size (be sure rims are not jagged)

PREPARATION

The paper and the cans can be prepared beforehand or prepared as part of the activity (see Procedure).

Although two cans with their tops completely removed can be used, the experiment will be more effective

(have fewer external effects), if only holes are placed in the cans’ lids, e.g., two holes from a bottle opener

to empty material out of the can, and one center hole created with an awl for the thermometer. Only one

hole is actually needed for the experiment - for the thermometer. Cans can optionally be filled with water,

or this can be done as a separate experiment to demonstrate the higher heat capacity of water compared to

air.

You can either use a flood lamp or a desk lamp (light bulb) to simulate sunlight, as described here, or

you can place the cans on a windowsill (window closed) or other sheltered area in direct sunlight. A flood

lamp will be the most effective option, causing the largest temperature increases.

PROCEDURE

Engagement

Discuss whether dark surfaces (e.g., asphalt) or bright surfaces (e.g., concrete) tend to get hotter in

sunlight. Which would you rather walk on during the day in the summertime? What color are solar cells,

for example those found on some calculators or freeway call boxes?

Explanation: I hope this helps! ∧    ∧

                                                ⊂∵→ω←∵⊃

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

For the work, applicate formula:

[tex]\boxed{\boxed{\green{\bf{W = F\times d}}}}[/tex]

According our data:

W = 12000 N * 1,5 m

W = 18000 J

The work done is 18000 Joules.

Answer:

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.

Explanation:

In one hour, the amount of sugar entering = 1000 kg

w.b moisture content is defined as,

weight of water / weight of water + weight of dry

[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100

[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering

it has 20% moisture content when entering

[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg

when leaving it has 3% moisture content then weight of dry material

[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg

[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03

[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03

[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800

[tex]W_{w} ^{'}[/tex] = 24.74 kg

When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.

Answer:

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.

Explanation: Given that

Maximum gradient = 500 m/km

Total distance = 1.5 km

Starting elevation = 20 m

Final elevation = 100 m

Gradient = change in elevation/ total distance.

Now, substitute the values into the formula.

Gradient = (100m - 20m)/1.5km

= 80m/1.5km

= 53.33m/km

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.

Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

What is the magnitude of its velocity just before it strikes the ground ?

The parameters given are:

Height H = 100m

Since the ball is thrown from a top of a building, initial velocity U = 0

Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

Substitute all the parameters into the formula

V^2 = 2 × 9.8 × 100

V^2 = 200 × 9.8

V^2 = 1960

V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

m is the mass

f is the force

From the question we have

[tex]a = \frac{126}{56.2} \\ = 2.241992...[/tex]

We have the final answer as

Hope this helps you

Answer:

D. 20J

Explanation:

Answer:

Explanation:

yes

Answer:

765,000Joules or 765kJ

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

Q = 765,000

Hence the amount of energy needed is 765,000Joules or 765kJ

Answer:

[tex]v=10m/sec[/tex]

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

[tex]F_N+Fg[/tex]

where

[tex]F_N =forceof the normal[/tex]

[tex]Fg= force due to gravity[/tex]

Generally the net force is given to be [tex]FC(force towards center)[/tex]

[tex]F_C =F_N + Fg[/tex]

[tex]F_N =F_C -Fg[/tex]

[tex]F_N=F_C-F_g[/tex]

[tex]F_N=\frac{mv^2}{R} -mg[/tex]

Mathematical we can now derive V

[tex]m_g + \frac{8m}{4}= \frac{mv^2}{8}[/tex]

[tex]\frac{5mg}{4} =\frac{mv^2}{8}[/tex]

[tex]v^2 =\frac{40*10}{4}[/tex]

[tex]v=10m/sec[/tex]

Therefore the speed of the roller coaster is given ton be [tex]v=10m/sec[/tex]

[tex]v=10m/s[/tex]

Given:

Radius of vertical r= 8m

To find:

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

[tex]F_N+F_g[/tex]

where

[tex]F_N=\text{Force of Normal}[/tex] and [tex]F_g=\text{Force due to gravity}[/tex]  

Generally the net force is given to be [tex]FC- \text{Force towards centre}[/tex]

[tex]F_C=F_N+F_g\\\\F_N=F_C+F_g\\\\F_N=\frac{mv^2}{r} -mg[/tex]

Mathematically we can now derive V

[tex]mg-\frac{8m}{4} =\frac{mv^2}{8}\\\\\frac{5mg}{4}=\frac{mv^2}{8}\\\\v^2=\frac{40*10}{4} \\\\v=10m/s[/tex]

Therefore, the speed of the roller coaster is given to be 10m/s.

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Answer:

Beat frequency = 125.5 Hz

Explanation:

Given:

Length of first organ pipes  = 43 cm = 0.43 m

Length of second organ pipes = 63 cm = 0.63 m

Computation:

Wavelength L = λ/2

length λ = 2L

Frequency f1 = v/λ = 340 / 2(0.43)

Frequency f1 = 395.34 Hz

Frequency f2 = v/λ2 = 340/2(0.63)

Frequency f2 = 269.84 Hz

Beat frequency  = Frequency f1 - Frequency f2

Beat frequency = 395.34 - 269.84

Beat frequency = 125.5 Hz

The question incomplete, the complete question is;

Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline.

(A)the one falling vertically

(B)the one on the incline

(C)Both have the same speed.

(D)cannot be determined

Answer:

(C)Both have the same speed.

Explanation:

When we consider the question closely, we will discover that an object falling down a frictionless incline is comparable to an object falling freely under gravity.

In both instances, the acceleration of objects is just the same irrespective of mass.

Hence, the object falling vertically and the object sliding down a frictionless plane will have the same speed at the bottom.

Answer:

122.5 m

Explanation:

From the question given above, the following data were obtained:

Time (t) = 5 s

Depth (h) =?

The depth of the well can be obtained as:

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

h = ½gt²

h = ½ × 9.8 × 5²

h = 4.9 × 25

h = 122.5 m

Thus, the depth of the well is 122.5 m