A ball is dropped out of a window and hits the ground at 14.5 m/s. How long did it take to fall to the ground?

Answers

Answer 1

Answer:

Explanation:

Use the one-dimensional equation:

[tex]v_f=v_0+at[/tex] which says that the final velocity of a falling object is equal to its initial velocity times the acceleration of gravity times the time it takes to fall. We have the final velocity, -14.5 (negative because its direction is down and down is negative), initial velocity is 0 (because it was held still by someone before it was dropped), and acceleration is -9.8 (negative again, because direction is down while acceleration increases). Filling in:

-14.5 = 0 - 9.8t and

-14.5 = -9.8t so

t = 1.5 seconds


Related Questions

Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.

Answers

Answer:

4.408 [tex]\mathsf{M_{sun}}[/tex]

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]

combined mass (m₁+m₂)  = 4.408 [tex]\mathsf{M_{sun}}[/tex]

By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects

Answers

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade

Answers

Answer:

The centripetal acceleration is 6.95 m/s²

Explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²



А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity

Answers

Thank lord for that please thank lord please thank

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 ss to speed up from rest to its top speed of 1 rotation every 1.30 ss . The astronaut is strapped into a seat 5.90 mm from the axis. What is the astronaut's tangential acceleration during the first 40.0 s?

How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.

Answers

Answer:

speed = 0.9 mm/s

Explanation:

time, t = 40 s

initial angular speed, wo = 0 rad/s

final frequency, f = 1/1.03 rps = 0.97 rps

final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s

time, t = 40 s

distance, r = 5.9 mm

The angular acceleration is given y the first equation of motion.

[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]

The linear velocity is

[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]

speed, v = 0.9 mm/s

A body of mass 5 kg is moved by a horizontal force of 0.5 N on a smooth frictionless table for 20 seconds. Calculate the change in kinetic energy.

A. 5 J

B. 20 J

C. 10 J

D. 30 J​

Answers

Answer: 10 J

Explanation:

Trust ;)

At what speed was object A moving ?

Answers

Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.

Answers

The answer is A because i know that it is and you will get the answer correct

If the loop is removed from the field region in a time interval of 2.8 ms , find the average emf that will be induced in the wire loop during the extraction process. Express your answer in volts.

Answers

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.

It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex]    and   [tex]$\phi_{B,f} = 0$[/tex]

Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]

  [tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]

  = 30.27 V

Therefore, the emf generated is 30.27 V.

 

The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N

Answers

Answer:

8760 N

Explanation:

think this is the right answer :)

A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?

Answers

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine

Answers

Fahrenheit scale is called the Rankine (°R) scale. These scales are related by the equations K = °C + 273.15, °R = °F + 459.67, and °R = 1.8 K. Zero in both the Kelvin and Rankine scales is at absolute zero.

is anyone online??just asking ​

Answers

Answer:

me...:(

Explanation:

Answer:

hello I'm online here thanks for the points (◔‿◔)

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?

Answers

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

Convert 387.1 K to °C

Answers

387.1 kelvins = 113.95 degrees celsius

a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.

PLEASE HELP I NEED THIS TODAY

Answers

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, what type of path will the particle follow

Answers

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

[tex]F = V\times B[/tex]

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?

a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved

Answers

(B) linear momentum of the system is conserved

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical speed be right before it hits the ground?
A. 0 m/s
B. 15 m/s
C. 40 m/s
D. 30 m/s

Answers

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 48 s. What is Jason's top speed

Answers

Answer:

81.1 m/s

Explanation:

The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration

So, T - f = ma

T -  μmg  = ma

a = T/m - μg

susbstμituting the values of the varμiables into the equation, we have

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 2.67 m/s² - 0.98 m/s²

a = 1.69 m/s²

Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s

So, v = u + at

v = 0 m/s + 1.69 m/s² × 48 s

v = 0 m/s + 81.12 m/s

v = 81.12 m/s

v ≅ 81.1 m/s

So, Jason's top speed is 81.1 m/s

A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter

Answers

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay​

Answers

Explanation:

4. Alin sa mga sumusunod na awitin ang may tempong presto?

a. “Chua-ay”

c. “Akong Manok”

b. “Sitsiritsit

d. "Ili-ili Tulog Anay”

5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?

a. mabagal

c. mabilis na mabilis

b. mabilis at mabagal d. katamtamang bilis

6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?

a. descant

b. ostinato

c. tempo

d. texture

7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

TANONG:

Kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay[tex] ?[/tex]SAGOT:

Maayos, Maganda, at iba pa...

If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N

Answers

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

What is Force?

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

Learn more about force.

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If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright

Answers

It will become 100 times less bright

Tarzan, whose mass is 75 kg, is running from a cheetah. Tarzan is moving at 5 m/s when he grabs onto a hanging vine. How high off the ground does Tarzan swing ?

Answers

Answer:

Explanation:

His Kinetic energy = 1/2 m v^2

v = 5 m/s

m = 75 kg

Ke = 1/2 75 * 5^2

Ke = 937.5 Joules

This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE

PE = m * g * h

m = 75

g = 9.81

h = ?

PE = 937.5

937.5 = 75 * 9.81 * h

937.5 = 735.75 * h

937.5/735.75 = h

h= 1.27 meters

A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.

Answers

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

Learn more hear:

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You’re working with a patient who suddenly falls. You should?

Answers

help them up and make sure there ok

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Answers

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kgg = 9.8 m/s²T = ?

Use formula:

[tex]\boxed{\bold{T=m*g}}[/tex]

Replace and solve:

[tex]\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{T=98\ N}}}[/tex]

The tension in the rope is 98 Newtons.

Greetings.

A system has both potential energy (PE) and kinetic energy (KE). According to
the law of conservation of energy, what can happen to the total energy of the
system?

Answers

Answer:

A. It must stay the same, but kinetic energy (KE) can be transformed to PE and PE can be transformed to KE within the system.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Potential energy (PE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

P.E represents potential energy measured in Joules.m represents the mass of an object. g represents acceleration due to gravity measured in meters per seconds square. h represents the height measured in meters.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where;

K.E represents kinetic energy measured in Joules. M represents mass measured in kilograms. V represents velocity measured in metres per seconds square.

Furthermore, the total energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Total energy = P.E + K.E

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

In this scenario, a system has both potential energy (PE) and kinetic energy (KE).

According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.

Which of these is a source of thermal energy inside earth

Answers

There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options

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