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3 years ago

A ball is thrown with an angle of 25.0 to the horizon with a speed of 18.0 m/s. What are its horizontal and vertical components?

1 answer:

6 0

Best Answer: <span> vi=24 m/s
θ=43°
x=16m

0) Use this equation because you have x component and not y.
You need time or part c so you solve it like this
x=(vi*cosθ)t
16=(24cos43°)t
t=.911552s=.912s

a) This is a trajectory and you need y so you use this equation.
y=(tanθ)x-(gx^2/(2(vi*cosθ)^2)
y=tan43°*16m-((-9.8*16m)^2)/(2(24*cos43°...
y=14.920m-39.901m
y=-24.9808m= -25.0m

b) Break it down into it's components.
vx=cosθ*vi
vx=cos43°*24
vx=17.5525m/s = 17.5m/s

c) vy= sin 43°*24-.5(-9.8)(.912s)
vy=22.2m/s

I'm not 100% positive on some of these but it should look something like the work above. </span>

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RACTIC PTUDIES

Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

3 0

3 years ago

Suppose that two point charges, each with a charge of +3.00 Coulomb are separated by a distance of 3.00 meters. Determine the ma

Sveta_85 [38]

Answer:

$9\cdot 10^9 N$

Explanation:

The magnitude of the electrical force between the two point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

$q_1 =q_2 = +3.0 C$ is the magnitude of each charge

r = 3.00 m is the separation between the two charges

Substituting the numbers into the formula, we find

$F=(9\cdot 10^9 Nm^2C^{-2})\frac{(+3 C)^2}{(3.00 m)^2}=9\cdot 10^9 N$

3 0

3 years ago

Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tent

luda_lava [24]

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

8 0

3 years ago

Read 2 more answers

Quick Quiz 40.2 While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: y

Firlakuza [10]

The order of decreasing photon energy is FM radio, AM radio, yellow light, and microwaves.

Electromagnetic radiation: The electromagnetic field's waves, which are conveying electromagnetic radiant energy through space, are what make up electromagnetic radiation. It consists of X-rays, gamma rays, microwaves, infrared, light, and radio waves. These waves are all a component of the electromagnetic spectrum.

Microwave radiation has wavelengths between one meter and one millimeter, which correspond to frequencies between 300 MHz and 300 GHz, respectively.

Radio waves: The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz and below.

The radio waves have the highest photon energy and the lowest is microwaves.

So, the highest to lowest order is FM radio, AM radio, yellow light and microwaves.

Learn more about waves here:

brainly.com/question/1968356

#SPJ4

4 0

1 year ago