A ball is launched from ground level at 30 m/s at an angle of 35° above the horizontal. how far does it go before it is at ground level again

Answers

Answer 1

Answer:

Explanation:

Ignoring air resistance

Initial vertical velocity is 30sin35 = 17.2 m/s

Gravity reduces this velocity to zero in a time of

t = v/g =17.2 / 9.8 = 1.755 s

it takes the same time to come back down to ground level for a total flight time of 2(1.755) = 3.51 s

The horizontal velocity is 30cos35 = 24.57 m/s

the distance traveled horizontally is

d = vt = 24.57(3.51) = 86.298... = 86 m


Related Questions

In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block

Answers

Answer:

Here we use the conservation of momentum theorem.

m stands for mass, and v stands for velocity. The numbers refer to the respective objects.

m1v1 + m2v2 = m1vf1 + m2vf2

Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.

m1v1 + m2v2 = vf(m1 + m2)

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. Find its density.

Answers

Answer:

Weight of object = 11.2 N

Apparent weight = 3.83 N     when immersed

Weight of water displaced = 11.2 - 3.83 = 7.37 N      

d (density) W / V       weight / volume      the weight density

Wo = Vo do    weight of object

Ww = Vo dw    where Ww is weight of equivalent volume of water = 7.37

Wo / Ww = do / dw    dividing previous equations

do = 11.2 / 7.37 dw = 1.52 dw

The density of the object is 1.52 that of water

The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3

So the weight density is 14900 N/m^3

An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.

To find the density, the given values are,

Weight in air = 11.20 N

Weight in water = 3.83 N

density of water = 1000 kg/m³

What is meant by Density?

According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.

Loss in weight of the object = Weight of object in air - weight of object in water

Loss in weight = 11.20 - 3.83 = 7.37 N

Volume of body x density of water x g = 7.37

Let V be the volume of body

V x 1000 x 9.8 =7.37

V = 7.5× 10⁻⁴ m³

Weight in air = Volume of body x density of body x g

11.20 = 7.5× 10⁻⁴ x d x 9.8

d = 1523 kg/m³.

Thus, the density of body is 1523 kg/m³.

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A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m

Answers

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

A car travels a certain distance from A to B with a speed of 60km/hr and then returns along the same path to the starting point with a speed of 40km/hr. Find the average speed and average velocity.
a) Km/hr
b) m/s

wrong answers will be reported!

Answers

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]

Answers

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid

Answers

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

What is density?

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?

Answers

Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

is it true that playing badmenton help you to become a better person?

Answers

Answer:

There is no scientific evidence that playing specifically l badminton makes you a better person, but sport and exercise in general release hormones which can make you feel more happy therefore making you nicer to the people around you and 'a better person'.

Answer:

It's true because playing any sport makes a person happy. So a happy person is a better person.

Please Mark as brainliest.

i just want an answer please

Answers

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

What is an ellipse?

a plane that slices between orbits

an oval-shaped orbit

a circular orbit

the center of gravity between orbiting objects​

Answers

Answer:

i think it's C thx correct me if wrong

Understanding what motivates anyone is not easy because each individual has different

Answers

Has different what????

Which of the following describes the motion of a block while it is in equilibrium? The block:
A. moves at a constant speed
B. slows down gradually to stop
C. speeds up for a bit, then moves at a constant speed
D. Accelerates constantly

Answers

The statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

EQUILIBRIUM:

A state of equilibrium in physics refers to a state of rest or the forces exerted on the object is in a balanced state.

In dynamic equilibrium, the acceleration of a body is zero. This means that the body is moving at a uniform speed.

Therefore, the statement that describes the motion of a block while it is in equilibrium is: The block moves at a constant speed.

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48.36
g.
MgSO4 to motes

Answers

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

Help me outtttt jejjejejeje

Answers

Answer:

do it got a picture

Explanation:

A cello and an organ are playing together. The organ plays a pitch of C with a frequency of 65.4 Hz in a pipe open at both ends. The cello plays its C string (a string fixed at both ends), and a beat frequency of 2.5 Hz is heard. The cellist loosens the string until no beat frequency is heard. what is the length of the pipe

Answers

Answer:

λ/2 = length of pipe

The pipe is open at both ends having wavelength of A-N-A or antinode-node-antinode which is 1/2 wavelength

λ = 331 m/s / (65.4 / s) = 5.06 m     wavelength of sound

L = 5.06 m / 2 = 2.53 m      length of pipe

The qualitative equivalent of external validity is:
A- Credibility

B- Dependability

C- Transformability

D- Confirmability

Answers

c transformability i think

in a compoumd are atoms physically or chemically combined

Answers

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.

Answers

Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]

What is the approximate value of k when 30 = e^5k?

Answers

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?

Answers

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must be true of the magnitude of the reaction force?

Answers

Newton's third law allows to find the result for the value of the reaction force during the collision is:

The reaction force is F = 90 N and is applied to the lighter body.

Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.

These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.

They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.

Consequently, the reaction force is F = 90 N directed towards the lighter body.

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A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?

What height will the football reach?

With what speed does the punter need to kick the football?

At what angle (θ), with the horizontal, does the punter need to kick the football?

Answers

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m

Answers

Answer:

Explanation:

f = [tex]\sqrt{T/(m/L)} / 2L[/tex]

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = [tex]\sqrt{120/12} /(2(3))[/tex])

f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

If the velocity and frequency of a wave are both doubled, how does the wavelength change?

Answers

The wavelength will remain unchanged.

Explanation:

The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is

[tex]v = \lambda\nu[/tex] (1)

so if we double both the velocity and the frequency, the equation above becomes

[tex]2v = \lambda(2\nu)[/tex] (2)

Solving for the wavelength from Eqn(2), we get

[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.​

Answers

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

a stone is thrown down off a bridge with a velocity of 22 m/s. what is its velocity after 1.5 seconds has passed?

Answers

Answer:

Velocity of the stone after 1.5 seconds has passed = 37 m/s

Explanation:

Initial velocity (u) = 22 m/s

Time (t) = 1.5 sec

Acceleration due to gravity (g) = 10 m/s²

By using kinematics equation:

v = u + gt

v = 22 + 10 × 1.5

v = 22 + 15

v = 37 m/s

Final velocity (v) = 37 m/s

Please help me.............................

Answers

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

A circular disk of radius 0.200 m rotates at a constant angular speed of 2.50 rev/s. What is the centripetal acceleration (in m/s2) of a point on the edge of the disk?

Answers

[tex]a_c = 3.14\:\text{m/s}^2[/tex]

Explanation:

First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:

[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]

The centripetal acceleration [tex]a_c[/tex] is defined as

[tex]a_c = \dfrac{v^2}{r}[/tex]

Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as

[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]

[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]

what type of stretching is beneficial for sports performance and involves momentum​

Answers

Answer:

Dynamic stretching

Explanation:

Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.

3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant velocity for 30s and come to rest with uniform acceleration of 2m/s2. Calculate the total distance covered by the car.​

Answers

The total distance traveled by the car at the given velocity and time is 900 m.

The given parameters:

initial velocity of the car, u = 20 m/sacceleration of the car, a = 12 m/s²time of motion of the car, t = 20 sfinal time = 30 sfinal acceleration = 2 m/s²

The final time of motion of car before coming to rest is calculated as follows;

[tex]v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s[/tex]

The graph of the car's motion is in the image uploaded.

The total distance traveled by the car is calculated as follows;

[tex]total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m[/tex]

Thus, the total distance traveled by the car at the given velocity and time is 900 m.

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