Answer:
B. Distance
Explanation:
When calculating speed, the value of the given distance is first entered on the calculator because it is in the numerator.
Speed is the rate of change of distance with time;
Speed = [tex]\frac{distance}{time}[/tex]
The value of the distance is inputted first before that of the time is entered.
This way the division sign evaluates for the speed.
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for the glove to reach its maximum height
Hello!!
For the maximum height the final velocity is zero, because can't up more.
Then, use the formula:
V = Vi + gt
Replacing, we have:
0 m/s = 5,3 m/s + (-9,8 m/s² * t)
0 m/s - 5,3 m/s = -9,8 m/s² * t
(-5,3 m/s) / -9,8 m/s² = t
t = 0,54 s
The time it will take to reach the maximum height is 0,54 seconds.
A food processor draws 8.47 A of current when connected to a potential difference of 110 V.
How much electrical energy is consumed by this food processor monthly (30 days) if it is used on average of 10.0 min every day?
Answer:
27.95[kW*min]
Explanation:
We must remember that the power can be determined by the product of the current by the voltage.
[tex]P=V*I[/tex]
where:
P = power [W]
V = voltage [volt]
I = amperage [Amp]
Now replacing:
[tex]P=110*8.47\\P=931.7[W][/tex]
Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]
[tex]Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min][/tex]
HELP PLZ!!!!!!!!!!!!!!
Answer:
Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)
Explanation:
In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)
For example, let's say you're trying to send the message "11001"
on on off off on
0 1 2 3 4 5 Numbers represent seconds
As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.
A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare
E=hf C=wavelength*F
E=hC/wavelength
E=(6.626*10^-34)*(3.00*10^8)/670*10^-9
E=(6.626*10^-34)*(3.00*10^8)/450*10^-9
A 430.0 g mass is being pushed by a 1.1 N net force. What is the acceleration?
Answer:
[tex]a = 2.56\ m/s^2[/tex]
Explanation:
Mechanical Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object.
The object has a mass of m=430 g = 0.43 Kg and is pushed by a net force of F=1.1 N. To calculate the acceleration, we solve for a:
[tex]\displaystyle a=\frac{F}{m}[/tex]
[tex]\displaystyle a=\frac{1.1}{0.43}[/tex]
[tex]\mathbf{a = 2.56\ m/s^2}[/tex]
1) A rock is thrown downward from the top of a 40.0-m-tall tower with an initial speed of 12 m/s. Assuming
negligible air resistance, what is the speed of the rock just before hitting the ground?
Answer:
By using [tex]v^{2}_{f} = v^{2}_{i} + 2a\Delta y[/tex], with [tex]v_{i} = -12m/s[/tex] and [tex]\Delta y = -40m[/tex]:
[tex]v^{2}_{f} = v^{2}_{i} + 2a/Delta y[/tex]
[tex]v^{2} = (-12m/s)^{2} + 2(-9.80m/s^{2})(-40m)[/tex]
[tex]v = -30m/s[/tex]
Explanation:
Hope this helped!
A rock is thrown downward from the top of a 40 m tall tower, with The speed of the rock just before hitting the ground will be equal to -30 m/s.
What is Friction?Friction is the resistance to a thing moving or rolling over another solid object. Although frictional forces can be advantageous, such as the traction required to walk while slipping, they also provide a significant amount of resistance to motion. In order to overcome frictional resistance in the moving parts, about 20% of an automobile's engine power is used.
The forces of attraction, also referred to as adhesion, between the contact zones of the surface, which are always minutely uneven, seem to be the main contributor to friction between metals.
From the given information in the question,
v²(f) = v²(i) + 2aΔy
v² = (-12 m/s)² + 2(-9.8)(-40)
v = -30 m/s.
Therefore, the velocity of the rock is -30 m/s.
To know more about Friction:
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If you exert 950 N*s of impulse on a 12 kg frictionless cart over the course of 5 seconds, how far will it travel during those seconds?
Answer:
The value is [tex]s =197.88 \ m[/tex]
Explanation:
From the question we are told that
The impulse is [tex]I = 950 \ N\cdot s[/tex]
The mass is [tex]m = 12 \ kg[/tex]
The time t = 5 s
Generally impulse is mathematically represented as
[tex]I = F * t[/tex]
=> [tex]F = \frac{I}{t}[/tex]
=> [tex]F = \frac{950 }{ 5 }[/tex]
=> [tex]F = 190 \ N[/tex]
Generally force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]a = \frac{F }{ m }[/tex]
=> [tex]a = \frac{190 }{ 12 }[/tex]
=> [tex]a = 15.83 \ m/s^2[/tex]
Generally from kinematic equation , the distance covered is
[tex]s = ut + \frac{1}{2} * at^2[/tex]
Here u is the initial velocity of the cart and the value is [tex]u = m/s[/tex]
=> [tex]s = 0 * 5 + \frac{1}{2} * 15.83 * 5^2[/tex]
=> [tex]s =197.88 \ m[/tex]
what belongs in the center section
Answer:
The second one I think
Explanation:
B
Pls Help physics 8th class question
Answer:
angle of reflection and angle of incident is always equal
Drag each item to indicate whether it is a characteristic of dark matter or baryonic matter. (2 points)
Dark Matter
Baryonic Matter
:: has been observed directly
:: interacts with baryonic matter
:: interacts with electromagnetic radiation
has not been observed directly
:: does not interact with electromagnetic radiation
++ does not interact with baryonic matter
Answer:
Dark matter:
- Doesn't interact with baryonic matter.
- It has not been observed directly
- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.
Explanation:
Dark matter:
- Doesn't interact with baryonic matter.
- It has not been observed directly
- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.
Baryonic matter:
- Has been observed directly because it includes nearly all the matter that we see in the world daily.
- It interacts with baryonic matter
- interacts with electromagnetic radiation
Dark Matter:
It has not been observed directly.It does not interact with baryonic matter.It does not interact with electromagnetic radiation.Baryonic Matter:
It has been observed directly.It interacts with baryonic matter.It interacts with electromagnetic radiation.Dark matter can be defined as often invisible substances that are difficult to spot because they don't absorb, emit or reflect light.
Hence, dark matter do not affect human view because they do not interact or interfere with electromagnetic radiation (force).
Although, humans can see right through the (weakly interacting) dark matter but it has not been observed directly.
Baryonic matter can be defined as a dark matter that is made up of baryons such as neutrons, and protons. Also, they are ordinary matter (both fermions and hadrons), as distinct from exotic forms.
In conclusion. baryonic matter has been observed directly and it can interact with electromagnetic radiation.
Read more: https://brainly.com/question/15677526
A toy car weighing 3.2 N gets pushed a distance of 10 m in 1.3 s. What is the momentum of the car?
Answer:
2.51 kg * m/s
Explanation:
In order to find momentum, use the equation below:
momentum = mass * velocity.
Since neither mass nor velocity was given, you must solve for both variables.
In order to solve for mass, use the force equation for its weight / gravitational force.
Fg (gravitational force) = 3.2 N = ma = 9.8m
mass = 3.2 N / 9.8 m/s^2 = 0.326531 kg
In order to solve for velocity, use the equation:
velocity = displacement / time
velocity = 10m / 1.3 s = 7.69231 m/s
Momentum = mass * velocity = 0.326531 kg * 7.69231 m/s = 2.51177 kg * m/s = 2.51 kg * m/s
A car accidently roll of a cliff. As it leaves the cliff it has horzontal velocity of 13 m/s it hits the ground 60m from the shoreline. Calculate the hight of the cliff
Answer: The height of the cliff is 104.59 m
Explanation:
The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.
Here we can use the relationship:
Time*Speed = Distance.
To find the time that the car is in the air, we know that:
speed = 13m/s
distance = 60m
time = T
13m/s*T = 60m
T = (60m)/13m/s = 4.62 s
This means that the car is falling for 4.62 seconds.
Now let's analyze the vertical problem.
As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero
The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:
g = 9.8m/s^2
then:
a = -9.8m/s^2
Where the negative sign is because the acceleration is pulling the car downwards.
To get the vertical velocity, we could integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:
v(t) = (-9.8m/s^2)*t
To get the vertical position equation, we need to integrate again over the time:
P(t) = (1/2)*(-9.8m/s^2)*t^2 + H
Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.
We know that the car needs 4.62 seconds to hit the ground, this means that:
P(4.6s) = 0m
Then:
P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0
(-4.9m/s^2)*(4.62s)^2 + H = 0
H = (4.9m/s^2)*(4.62s)^2 = 104.59 m
This means that the cliff is 104.59 meters high
A sound wave enters a new medium where sound travels faster. How does this affect the frequency and wavelength of the sound?
А. The frequency increases and the wavelength decreases.
B. The frequency decreases and the wavelength increases.
C. The frequency stays the same and the wavelength increases.
D.The frequency stays the same and the wavelength decreases.
E. Neither the frequency nor the wavelength is affected.
Answer:
The frequency stays the same and the wavelength decreases.
Explanation:
When sound wave enters a new medium where sound travels faster, its frequency will remain same because it depends only on the source.
The relation between wavelength and speed is inverse, it means when the speed of sound increases, its wavelength will decrease.
So, the frequency stays the same and the wavelength decreases. Hence, the correct option is (d).
A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.
Answer:
330
Explanation:
(a) The torque the net thrust produces about the center of the circle is of 20 N-m.
(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Given data:
The mass of model airplane is, m = 1.0 kg.
The radius of horizontal circle is, r = 20.0 m.
The magnitude of net thrust by engine is, F = 1.0 N.
(a)
The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,
T = F × r
Solving as,
T = 1.0 × 20.0
T = 20 N-m
Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.
(b)
The expression for the angular acceleration of airplane during the horizontal flight is given as,
[tex]T = I \times \alpha[/tex]
Here, I is the moment of inertia of airplane and its value is,
[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]
So, the angular acceleration is,
20 = 200 × α
α = 20/200
α = 0.1 rad/s²
Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Learn more about the torque here:
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An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.
Answer:
The answer is "0.5555 m"
Explanation:
Where the reference leaves the list and the viewer is at rest:
[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]
[tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]
[tex]=0.5555 \ m[/tex]