Answer:
The car takes 7 seconds to speed up and get out of the way
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes uniformly over time.
Having a as the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
v_f=v_o+at
Using the above equation we can solve for t:
[tex]\displaystyle t=\frac{v_f-v_o}{a}[/tex]
The car has an initial speed of vo=25 m/s and ends up at vf=60 m/s with an acceleration of a=5 [tex]m/s^2[/tex].
Given all the appropriate data, calculate t:
[tex]\displaystyle t=\frac{60-25}{5}[/tex]
Calculating:
[tex]\displaystyle t=\frac{35}{5}[/tex]
t = 7 seconds
The car takes 7 seconds to speed up and get out of the way
ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.
In working this problem, assume the value of "g" to be 9.8 m/s2 with two (2) significant digits unless otherwise stated.
A machine exerts a 100 N force to the right over a 5.00 meter length in 4.00 seconds. Calculate the power output of this machine.
_____ W
500.
125
80.0
0.200
Answer:
The answer is 500
Explanation:
Define Gulf in your own words?
A constant force pushes a 5 kg brick at 10 N. If the mass was tripled, what would happen to the acceleration if the same force is applied?
(PLEASE SHOW WITH STEPS)
Explanation:
The original acceleration is:
∑F = ma
10 N = (5 kg) a
a = 2 m/s²
If the mass is tripled, the new acceleration is:
∑F = ma
10 N = (15 kg) a
a = 0.67 m/s²
The acceleration is reduced by a factor of 3.
Answer:
1.35 miles per second
Explanation:
1 step it depende the way and hight he is
2 step the mass of the brick it depends
3 if it was triplet the acceleration it would be 1.50
Question 2 of 10
Which of the following careers would physicists be least qualified to perform,
based on their training?
A. Nuclear engineer
B. Solar panel designer
C. Research lab manager
D. Gardener
Answer:
a and d
Explanation:
a because a nuclear engineer is more on explosions which links to chemistry and d because Gardner is irrelevant
29. Identify whether each of the following is a form of mechanical energy:
elastic potential energy, chemical potential energy, gravitational potential
energy.
Explanation:
elastic potential energy, chemical potential energy, and gravitational potential energy are all forms of mechanical energy.
Mechanical energy includes elastic potential energy, chemical potential energy, and gravitational potential energy.
What is Mechanical energy?Mechanical energy is the energy that an object has as a result of its motion or position. Because it is stretched out in that position, a pulled bow has mechanical energy (elastic potential energy). Chemical potential energy is the energy that a substance has stored in its chemical bonds. A barbell hoisted high above a weightlifter's head has mechanical energy because of its vertical position above the ground (gravitational potential energy).The sum of kinetic and potential energy in an object used to perform a specific task is known as mechanical energy. In other words, it describes the energy of an object as a result of its motion, position, or both.
To learn more about mechanical energy, refer
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Bird man is flying horizontally at a speed of 33m/s and a height of 86m. Bird man releases a turd directly above the start of the field. How far from the start of the field should the robot hold the bucket to catch the turd.
Answer:
2838
Explanation:
20 points HELP‼️ Which is more important in developing who
you are-environment or genetics? Explain why you believe this.
Answer:
Environment
Explanation:
Your Environment, More Than Genetics, Determines Your Immune Health. When it comes to immunity, the environment you grow up in, or how you were 'nurtured' , is more important than nature, a new study suggests. Particularly as you get older.
is it true or false that chemical change like baking a cake normally has a color change?
the inside surfaces of the glass inside thermos bottles are coated with aluminum to prevent energy transfers by
Answer:
radiation heat.
Explanation:
That is any heat that passes through the air, like the sun.
Answer:
radiation
Explanation: I’m sure
pls mark as brainliest if possible
4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2° below the horizontal. Find mk between the box and the floor.
Answer:
0.61°
Explanation:
Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.
Pulling force= resistance force
From the formula for pulling force,
F(x)= Fcos(θ)
= 425×cos(35.2)
=347N
The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)
=425×0.567=245N
Resistance force= (325N+ 245N) (α)= 570N(α)
We can now equates the pulling force to resistance force
570 (α)= 347N
(α)= 347/570
= 0.61
What is the mass of an object on the moon whose weight sitting on the Earth is 1900 N?
Reena has rolled down her toy car through a ramp. When she increases the height of the ramp (10 cm ) the car has travelled 5 metres and she increases the height of the ramp (13 cm) the car has travelled 6 metres . What happened when she raises the height of the ramp?
Answer:
As she raises the height of the ramp, the speed of the toy car decreases.
Explanation:
Speed is the ratio of the distance traveled by an object to the time taken.
i.e speed = [tex]\frac{distance}{time}[/tex]
From the given question, it would be observed that:
i. when the height of the ramp was 10 cm, the car traveled 5 meters.
ii. when the height of the ramp was increased to 13 cm, the car traveled 6 meters.
Therefore, an increase in the height of the ramp causes a decrease in the speed of the toy car so that the distance traveled per unit time increases. Showing that there is an inverse relationship between the height of the ramp and the speed of the toy car.
A juggler throws a beanbag into the air with a speed of 1.0 m/s.
How long does it take for the beanbag to reach its maximum height?
41. A statue weighs 1,000N and exerts a pressure of 20,000 Pa. How big is
the base of the statue in square meters?
please help
Answer:
The answer is 0.05 m²Explanation:
The area of the base of the statue can be found by using the formula
[tex]a = \frac{f}{p} \\ [/tex]
f is the force
p is the pressure
From the question we have
[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]
We have the final answer as
0.05 m²Hope this helps you
gravitational forces is ------ force
Answer:
Gravitational force is noncontact force
Explanation:
Contact force occurs due to the contact between two different objects. Non-contact force occurs due to either attraction or repulsion between two objects such that there is no contact between these objects. There is no field linked with the contact force. ... Gravitational force is an example of a non-contact force.
based on your reading "take a closer look" what can you say about the image that forms on the retina of your eye?
Answer:
it is an upside-down image but the brain realizes it and corrects it in its orientation so we see it in right direction.
Answer:
The lens of the eye is a convex lens. The image that it forms on the retina is upside down.
Explanation:
The eye makes an image, but it is upside down, the brain corrects the image and turns it right side up
Which statement accurately describes planetesimals?
A. They are the origins of planets.
B. They formed from ice and rocks.
C. They were created during the big bang.
D. They contain 98% of matter in the solar system
Answer:
B
Explanation:
mark me as brainlist
An archer stands on the ground and fires an arrow at a target. A second archer stands at the top of a building and holds an arrow in his hand. Which arrow has more potential energy?
In the same scenario described in question 7, which arrow has more kinetic energy? Explain.
Answer:
The arrow of the second archer standing on a high building will have more potential energy.
The arrow of the first archer standing on ground will have more kinetic energy.
Explanation:
POTENTIAL ENERGY:
The potential energy of an object depends upon its height, as given in the formula:
P.E = mgh
Hence, the arrow with the greater height will have more potential energy.
Therefore, the arrow of the second archer standing on a high building will have more potential energy.
KINETIC ENERGY:
The kinetic energy of an object depends upon its speed, as given in the formula:
K.E = (1/2)mv²
Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.
Therefore, the arrow of the first archer standing on ground will have more kinetic energy.
(a) The second archer standing at the top of the building will have more potential energy.
(b) The first archer on the ground level will have the more kinetic energy.
The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.
P.E = mgh
where;
h is the height above the groundThus, the second archer standing at the top of the building will have more potential energy.
(b) The kinetic energy of an object is the energy possessed by the object due to its motion.
K.E = ¹/₂mv²
where;
v is the speed of the objectAn object in motion, has zero velocity at maximum height but maximum velocity on the ground level.
Thus, the first archer on the ground level will have the more kinetic energy.
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HELP HURRY PLEASE
According to the law of reflection, if angle of incidence is 50, what is angle of reflection?
A. 45
B. 50
C. 25
D. 100
Answer:
50
Explanation:
The angle of reflection is 50 because angle of incidence is equal to the angle of reflection
Which best compares kinetic energy and temperature?
Answer:
^ correct
Explanation:
Option A on edge 2020
What current flows through a bulb if 360 C of charge moves through
a bulb in 20 minutes?
Answer: A current of 1 A is flowing in a circuit if a charge of 1 coulomb passes any point in the circuit every second.
1 Amp = 1 Coulomb per second
We can write this formula as:
Current (I) = Charge (Q) / Time (t)
Charge (Q) = Current (I) x Time (t)
Explanation:
A rocket is launched from the origin with an acceleration of 20.0 m/s2 in a straight line at an angle of 30.0 degrees above the horizontal. The launch acceleration lasts for 2.00 seconds at which time the fuel is exhausted. The rocket then falls with an acceleration of 9.80 m/s2 downward. What is the time it takes to reach maximum height?
Answer:
4.04 seconds
Explanation:
Context before solving:In order to solve this problem, we must keep in mind that the initial 2.00 seconds at which the fuel is exhausted does not signal when the rocket reaches its maximum height.
From this moment on, the rocket has a downward acceleration of 9.8 m/s² but it still has an upwards velocity, which we will calculate. This upwards velocity keeps the rocket moving up for a certain period of time, which we will also calculate.
For this problem, let's set the upwards direction to be positive and the downwards direction to be negative.
To find the time that the rocket takes to reach its maximum height, we are going to use the initial 2.00 seconds and find the additional seconds it takes to reach a final vertical velocity of 0 m/s. This represents the time at which the rocket stops moving and heads in the downwards direction, which takes place after the rocket has reached its maximum height.
Solving for initial velocity:The time in the air of an object in projectile motion can be found using this equation, derived from one of the constant acceleration kinematic equations.
Time in the air (projectile motion):
[tex]$t=\frac{2v_isin\theta}{g}[/tex] where g = gravitation acceleration = 9.8 m/s²Solve for [tex]v_i[/tex] by plugging in 2.00 seconds for t, 30 degrees for theta, and 20.0 m/s² for acceleration, since this is not a constant acceleration problem.
[tex]$2.00=\frac{2v_isin(30)}{20.0}[/tex]Multiply sin(30) and 2 together.
[tex]$2=\frac{v_i}{20}[/tex]Multiply 20 to both sides of the equation.
[tex]$40=v_i[/tex] [tex]v_i=40[/tex]Finding the vertical component:Now we know that the initial velocity of the rocket is 40 m/s. We need to solve for the vertical component of the rocket's velocity in order to solve for the additional time it took after the 2.00 seconds to reach its maximum height.
Vertical component:
[tex](v_i)_y=v_i \times sin\theta[/tex] [tex](v_i)_y=(40) \times sin(30)[/tex][tex](v_i)_y=20[/tex]The vertical component of the velocity vector is 20 m/s.
Finding additional seconds after 2.00 s:Now, in order to solve for the additional seconds that the rocket took to reach its maximum height, let's use one of the kinematic constant acceleration equations that uses the variables [tex]v_f[/tex], [tex]v_i[/tex], [tex]a[/tex], and [tex]t[/tex].
[tex]v_f=v_i + at[/tex]Since we are trying to solve for time, we need to use this equation in terms of the vertical direction, aka the y-direction. Time is the same in either case.
[tex](v_f)_y=(v_i)_y+a_yt[/tex]The final vertical velocity of this rocket is 0 m/s at the top, or its maximum height. We found that the vertical component, aka the rocket's initial vertical velocity, is 20 m/s. The acceleration is given to us: -9.8 m/s² (since it's falling downwards, the acceleration must be negative because we already established this in the beginning).
We are trying to solve for time t. Substitute the known values into the equation.
[tex]0=(20)+(-9.8)t[/tex]Subtract 20 from both sides of the equation.
[tex]-20=-9.8t[/tex]Divide both sides of the equation by -9.8.
[tex]2.040816327=t[/tex] [tex]t=2.04\ \text{seconds}[/tex] Finding total time to reach max height:Now we can take this time and add it to the initial 2.00 seconds of the rocket. This time we just solved for is the time after these initial seconds that the rocket kept going upwards, since its initial vertical velocity was not 0 m/s yet.
[tex]2.00+2.04=4.04 \ \text{seconds}[/tex]The time that the rocket takes to reach its maximum height is 4.04 seconds.
Answer:
4.04 second is your answer hope it help you........
The components of a 15-meters-per-second velocity at an angle of 60.° above the horizontal are (1) 7.5 m/s vertical and 13 m/s horizontal (2) 13 m/s vertical and 7.5 m/s horizontal (3) 6.0 m/s vertical and 9.0 m/s horizontal (4) 9.0 m/s vertical and 6.0 m/s horizontal
Answer:
13 m/s vertical and 7.5 m/s horizontal
Explanation:
❤️
The vertical and horizontal components of a projectile is required.
The correct option is (2) 13 m/s vertical and 7.5 m/s horizontal
u = Initial velocity of projectile = 15 m/s
[tex]\theta[/tex] = Angle of projectile = [tex]60^{\circ}[/tex]
The horizontal component
[tex]u_x=u\cos \theta\\\Rightarrow u_x=15\cos60^{\circ}\\\Rightarrow u_x=7.5\ \text{m/s}[/tex]
The vertical component
[tex]u_y=u\sin \theta\\\Rightarrow u_x=15\sin60^{\circ}\\\Rightarrow u_x=13\ \text{m/s}[/tex]
Learn more:
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A car accelerates uniformly from rest to a final speed of 19 m/s in 10. seconds. How far does it travel during this period of acceleration?
Answer:
x = 95 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f} =v_{o} +a*t[/tex]
where:
Vf = final velocity = 19 [m/s]
Vo = initial velocity = 0 (starts from the rest)
a = acceleration [m/s²]
t = time = 10 [s]
Now we can find the acceleration
19 = 0 *a*(10)
a = 1.9 [m/s]
With the second equation we can find the distance:
[tex]v_{f} ^{2} =v_{o} ^{2} +(2*a*x)[/tex]
where:
x = distance [m]
(19)² = (0)² + (2*1.9*x)
3.8*x = 361
x = 95 [m]
A cannon sitting on the edge of a cliff 500 meters above the ground shoots a cannon ball with a horizontal velocity of 100m/s. At some point the cannon hits the ground. How far away from the point it was launched will the cannon ball land?
Answer:
1010 m
Explanation:
The following data were obtained from the question:
Height (h) of cliff = 500 m
Horizontal velocity (u) = 100 m/s
Horizontal distance (s) =?
Next we shall determine the time taken for the cannon ball to hit the ground. This can be obtained as follow:
Height (h) of cliff = 500 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
500 = ½ × 9.8 × t²
500 = 4.9 × t²
Divide both side by 4.9
t² = 500/4.9
Take the square root of both side
t = √(500/4.9)
t = 10.1 s
Finally, we shall determine the horizontal distance travelled by the cannon ball.
This can be obtained by using the following formula (s = ut) as illustrated below:
Horizontal velocity (u) = 100 m/s
Time (t) = 10.1 s
Horizontal distance (s) =?
s = ut
s = 100 × 10.1
s = 1010 m
Thus, the cannon ball was launched 1010 m away from the cliff.
Please help I need this right now
Answer:
The basket ball
Explanation:
The heavier an object is the more kinetic energy it will have even if it is thrown at the same speed; therefore, since the basketball is the heaviest, it is the logical answer.
While a marble is heavy for its size, it would not have the same kinetic energy as the basket ball.
Beach balls weigh the least out of the options given so this cannot be the answer.
Golf balls, while they are dense, they do not weigh nearly as much as a basketball would!
Hope this helps <3
Answer:
The answer is B the basketball
Explanation:
is it true that scientific theories and scientific laws both describe observed events??? provide evidence!!!!! only 2 people could answer so if you can't then comment and provide evidence thank you very much!!!!!!
Answer:
I'm not positive but I think the answer is Yes, because both scientific laws and theories are based on evidence.
Explanation:
Sorry I don't have notes on this, hope this helps! :)
Which device consists of two separate coils of wire wrapped around an iron
ring?
O A. A transformer
B. A generator
C. A motor
D. A commutator
Answer:
It would be transformer
Explanation:
hope this helped
Answer:
transformer
Explanation:
got it right
To get to the kitchen from her bedroom, Maria walks 3 m north and then 4
m east. What is the total distance and displacement of Maria? *
A student determines the density ρ of steel by taking measurements from a steel wire
Mass- 6.2 +-0.1g
Length- 25.0 +-0.1m
Diameter- 2.00 +-0.01mm
He uses the equation ρ= 4m/πd^2l
What is the percentage uncertainty in his calculated value of density ?
Answer:
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
Explanation:
We can estimate the absolute uncertainty by the definition of total differential. That is:
[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)
Where:
[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].
[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.
[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.
[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.
[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.
[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.
[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.
Partial derivatives are, respectively:
[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)
[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)
[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)
And we expand (1) as follows:
[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]
[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)
If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:
[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]
[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]
And the expected density is:
[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)
[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]
[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]
The percentage uncertainty in his calculated value of density is:
[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)
If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:
[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]
[tex]\%e = \pm 0.713\,\%[/tex]
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].